电子邮件附件中的 .p7s是什么文件,是怎么来的?哪位大神可以详细介绍下

使用outlook的签名功能发送邮件,在接收方可以看到附件中有个 .p7s的文件。这个是什么文件?是怎么来的?怎么解析?

1个回答

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怎么得到以下原始二进制字符串的真实内容?
请问以下这个原始字符串这个是通过什么编码的呢?我用assii解密后得到后面的assii格式的字符串,但是也是一堆东西,只有9000000360是明文数据 # 原始字符串: 0060100030303030304e003000000000393030303030303336300000000000000e0707070707070b0e083838383831395f5f080d0904020d3a3a3a3a3333333b3b3b3b343434343c3c3c35353535353d3d36360f0f1806100719020f3737373838383838383831393939393909021f0f0a170f0c1e7207160c0216767234343c 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 000a100130303030304e003000000000393030303030303336300000000000000952030c1d5e5b011b62 0040100530303030304e003000000000393030303030303336300000000000000052030c1d5e5b021c6334614633213a05520c1a5f53021c543562472d223b64353e2f6966332c553663412e233c65363f296a57041d66075d422f243d663739 0000100530303030304e00300000000039303030303030333630000000000000 0040101230303030304e003000000000393030303030303336300000000000000052030c1d5e5b021c6334614633213a05520c1a5f53021c543562472d223b64353e2f6966332c553663412e233c65363f296a57041d66075d422f243d663739 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 002b123330303030304e003000000000393030303030303336300000000000006d307278683c2275176a04517603110a50020d2f3b37736f0462267a4b440a5005081f5e677763127f277c 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 002d124030303030304e003000000000393030303030303336300000000000006d307278683c2275176a04517603110a50020d2f3b37736f0462267a4b440a5005081f5e677763127f277c2f22 00c3124030303030304e00300000000039303030303030333630000000000000602f7c7268376a78641e7b2f036a20753660706121206b2a1267270277677f297a706a30677b6319722f0e7d703d20787e6b26227962187204436a6a7c247b7c653e3c7861197407427d6a702268686e21287f61507c1a137c676d2a776f7a2c26366f1b612d0c762678316079793d38646755055075070f0252350d005f55331b7a0552461d0c0b54340c1850480b18540752721a0d0454370e1859531a1567366c731c10135e0638195f5c1b166330677300150e5701091d5c68071b660b71720420 003b101130303030304e003000000000393030303030303336300000000000000052030c1d5e5b021c6334614633213a05520c1a5f53021c543562472d223b64353e2f6966332c553663412e233c65363f296a56052d56375d422f 0000101130303030304e00300000000039303030303030333630000000000000 # # ACSII解码后的字符串: .`..00000N.0....9000000360................888819__. .. ::::333;;;;4444<<<55555==66.........7778888888199999 ... ....r.....vr44<. ..00000N.01...9000000360........V...^] .i ey.9. .. ........... ...... .............. .................................. .4...................7.......8.....................3; ..H... @....I....66...............................Rj.J^. ...... ...5..............ny...... ........... ::333;;;;4444<<<. ..00000N.0....9000000360...... R...^[..b.@..00000N.0....9000000360.......R...^[..c4aF3!:.R.._S..T5bG-";d5>/if3,U6cA.#<e6?)jW..f.]B/$=f79....00000N.0....9000000360.......@..00000N.0....9000000360.......R...^[..c4aF3!:.R.._S..T5bG-";d5>/if3,U6cA.#<e6?)jW..f.]B/$=f79.3..00000N.0....9000000360.......R...^[..c.Af.........2.t.Bg ..D...IF..u.Ca...E.. JG. v.}b...F.. KH..w.~c...G...LI..q..d...A...MJ. r.@e...B.. NK. s./.}otC...OE..t.RG ..D...iF..u.Cq...E.. JG. v.}b...R...KH..w.~c...W. .LI..q..d...Q...MJ. r.@e...S...NK. s.Af...C...OE..t.Bg ..D...IF..u.Ca...E...JG. v.}b...F...kh5.W1^C0%>g1:+li6/Q2_D1&?a2;,mj.+.300000N.0....9000000360......m0rxh<"u.j.Qv.. P. /;7so.b&zKD P...^gwc..'|.!.300000N.0....9000000360......n#ayd*jfk.q/.~d;4uok= .j.{c.llh*`s.*)`x.d*.k"r ao{#$q,.v. jpk'{8b$$qb.b Bviq&ejn>*sa.3..xrm$bzu#?z(.r).`bi#jqb .h+.{3.voo.5{{#)|j.pc.ccy(pnf*-}y.7".}fh6ejdk {b.d. aadgqko.2p`.|_.yws.u<.\X..`3............(c2QD...S ..oX..}.UG...M. /]R..l.cs...] ?.hV..e7S@......([f..a.st.%.S. .jX4.g.ns.. 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两个JS文件冲突!只要有hover()事件的js和底下的js文件同时用就报错!
eval(function (p, a, c, k, e, d) { e = function (c) { return (c < a ? '' : e(parseInt(c / a))) + ((c = c % a) > 35 ? 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0, {}))
Mahjong 问题的求解
Problem Description Japanese Mahjong is a four-player game. The game needs four people to sit around a desk and play with a set of Mahjong tiles. A set of Mahjong tiles contains four copies of the tiles described next: One to nine Man, which we use 1m to 9m to represent; One to nine Sou, which we use 1s to 9s to represent; One to nine Pin, which we use 1p to 9p to represent; Character tiles, which are:Ton, Nan, Sei, Pei, Haku, Hatsu, Chun, which we use 1c to 7c to represent. A winning state means a set of 14 tiles that normally contains a pair of same tiles (which we call "eyes") and four melds. A meld is formed by either three same tiles(1m, 1m, 1m or 2c, 2c, 2c for example) or three continuous non-character tiles(1m, 2m, 3m or 5s, 6s, 7s for example). However, there are two special winning states that are different with the description above, which are: "Chii Toitsu", which means 7 different pairs of tiles; "Kokushi Muso", which means a set of tiles that contains all these tiles: 1m, 9m, 1p, 9p, 1s, 9s and all 7 character tiles. And the rest tile should also be one of the 13 tiles above. And the game starts with four players receiving 13 tiles. In each round every player must draw one tile from the deck one by one. If he reaches a winning state with these 14 tiles, he can say "Tsu Mo" and win the game. Otherwise he should discard one of his 14 tiles. And if the tile he throws out can form a winning state with the 13 tiles of any other player, the player can say "Ron" and win the game. Now the question is, given the 13 tiles you have, does there exist any tiles that can form a winning state with your tiles? (Notes: Some of the pictures and descriptions above come from Wikipedia.) Input The input data begins with a integer T(1≤T≤20000). Next are T cases, each of which contains 13 tiles. The description of every tile is as above. Output For each cases, if there actually exists some tiles that can form a winning state with the 13 tiles given, print the number first and then print all those tiles in order as the description order of tiles above. Otherwise print a line "Nooten"(without quotation marks). Sample Input 2 1s 2s 3s 2c 2c 2c 2p 3p 5m 6m 7m 1p 1p 1p 1p 2p 3p 4s 5s 6s 7c 7c 3s 3s 2m 2m Sample Output 2 1p 4p Nooten
mongodb 为什么高版本api写入比低版本api写入效率低
* 同一台服务器上,非集群单机测试,插入2个共1.1GB数据 * 写入数据用的java高版本api是3.8,低版本api是2.4 * mongo2.4不支持高版本api --- ![图片说明](https://img-ask.csdn.net/upload/201912/06/1575621736_346475.png) --- 用低版本2.4api的collection.insert(listdbo);每提交10万行耗时4s 用高版本3.8api的collection.insertMany(listdbo);每提交10万行耗时7s 低版本api写入mongodb2.4和mongodb4.0耗时基本一样 --- # **为什么高版本api写入mongodb4.0耗时会比较长????**
Mahjong 的问题
Problem Description Japanese Mahjong is a four-player game. The game needs four people to sit around a desk and play with a set of Mahjong tiles. A set of Mahjong tiles contains four copies of the tiles described next: One to nine Man, which we use 1m to 9m to represent; One to nine Sou, which we use 1s to 9s to represent; One to nine Pin, which we use 1p to 9p to represent; Character tiles, which are:Ton, Nan, Sei, Pei, Haku, Hatsu, Chun, which we use 1c to 7c to represent. A winning state means a set of 14 tiles that normally contains a pair of same tiles (which we call "eyes") and four melds. A meld is formed by either three same tiles(1m, 1m, 1m or 2c, 2c, 2c for example) or three continuous non-character tiles(1m, 2m, 3m or 5s, 6s, 7s for example). However, there are two special winning states that are different with the description above, which are: "Chii Toitsu", which means 7 different pairs of tiles; "Kokushi Muso", which means a set of tiles that contains all these tiles: 1m, 9m, 1p, 9p, 1s, 9s and all 7 character tiles. And the rest tile should also be one of the 13 tiles above. And the game starts with four players receiving 13 tiles. In each round every player must draw one tile from the deck one by one. If he reaches a winning state with these 14 tiles, he can say "Tsu Mo" and win the game. Otherwise he should discard one of his 14 tiles. And if the tile he throws out can form a winning state with the 13 tiles of any other player, the player can say "Ron" and win the game. Now the question is, given the 13 tiles you have, does there exist any tiles that can form a winning state with your tiles? (Notes: Some of the pictures and descriptions above come from Wikipedia.) Input The input data begins with a integer T(1≤T≤20000). Next are T cases, each of which contains 13 tiles. The description of every tile is as above. Output For each cases, if there actually exists some tiles that can form a winning state with the 13 tiles given, print the number first and then print all those tiles in order as the description order of tiles above. Otherwise print a line "Nooten"(without quotation marks). Sample Input 2 1s 2s 3s 2c 2c 2c 2p 3p 5m 6m 7m 1p 1p 1p 1p 2p 3p 4s 5s 6s 7c 7c 3s 3s 2m 2m Sample Output 2 1p 4p Nooten
at91sam7s64对外部时钟计数时计数值始终为0
at91sam7s64的TC有5个内部时钟源和3个外部时钟可以选择,并且在时钟上升沿计数器加1,现在选择TC1对外部信号计数,外部信号连接到PA26,将PA26复用到外设B即TIOA2,然后选择其作为时钟,且这个引脚是有高低电平变化的,配置的寄存器如下: *AT91C_PIOA_PDR = 1<<26 ; //PA26禁止IO 使能外设 *AT91C_PIOA_BSR = 1<<26 ; //PA26分配给外设B即TIOA2 PA15分配给外设B即TIOA1 *AT91C_PMC_PCER = 1<<AT91C_ID_PIOA; //使能IO控制器时钟 需要吗? P204说明 *AT91C_TCB_BMR = AT91C_TCB_TC1XC1S_TIOA2 ; //TC_BMR中设置XC1的时钟为TIOA2 *AT91C_PMC_PCER = 1<<AT91C_ID_TC1; //允许TC1口的时钟 *AT91C_TC1_CCR = AT91C_TC_CLKEN; //使能计数时钟 *AT91C_TC1_CMR = AT91C_TC_CLKS_XC1; //设置时钟为XC1 *AT91C_PIOA_ODR = 1<<26 ; //输出禁用 *AT91C_PIOA_PPUDR = 1<<26 ; //上拉电阻禁用 改为PUER的话是使能 *AT91C_TC1_CCR = 0x04; //计数器复位并启动时钟 然后读取*AT91C_TC1_CV的值始终为0,这是哪里不对了?还有就是把*AT91C_TC1_CMR = AT91C_TC_CLKS_XC1; 这句改为选择内部时钟是可以计数的,但是换成外部的XC1就不能,求大神指点
Xcode armv7s的错误。。。。。
Undefined symbols for architecture armv7: "_GDLocalizedString", referenced from: -[QBAssetCollectionViewController done] in QBAssetCollectionViewController.o -[QBAssetCollectionViewController assetCell:canSelectAssetAtIndex:] in QBAssetCollectionViewController.o -[QBImagePickerController initWithNibName:bundle:] in QBImagePickerController.o ld: symbol(s) not found for architecture armv7 clang: error: linker command failed with exit code 1 (use -v to see invocation) 这个怎么解决啊,求大神指点
函数scoresort要求按学生平均成绩从高到低排序成绩表,为什么冒泡法和选择法都无法正确排序是逻辑错误吗?
``` #include<stdio.h> #include<string.h> struct student { long int num; char name[8]; double score[3]; }; struct arrayst { struct student stu; double nums; }; int main() { void input(struct arrayst a[],int x); void output(struct arrayst arr[],int x); void scoresort(struct arrayst arr[],int x); arrayst a[100]; int n; while(1) { printf("请输入一个在5到100之间的数:"); scanf("%d",&n); if(n>=5 && n<=100) break; } printf("NUM MAME SCORE1 SORE2 SCORE3\n"); input(a,n);//建立学生成绩表 printf("Before sort:\n"); printf("NUM MAME SCORE1 SORE2 SCORE3 AVERAGE\n"); output(a,n);//输出学生成绩表 scoresort(a,n);//按平均成绩从高到低排序学生成绩表 printf("After Sort:\n"); printf("NUM MAME SCORE1 SORE2 SCORE3 AVERAGE\n"); output(a,n);//输出学生成绩表 return 0; } void input(struct arrayst a[],int x) { int i; for(i=0;i<x;i++) { scanf("%d%s%lf%lf%lf",&a[i].stu.num,&a[i].stu.name,&a[i].stu.score[0],&a[i].stu.score[1],&a[i].stu.score[2]); } } void output(struct arrayst arr[],int x) { float aver; int i; for(i=0;i<x;i++) { aver=(arr[i].stu.score[0]+arr[i].stu.score[1]+arr[i].stu.score[2])/3.0; printf("%-6d%-7s%-9.0lf%-8.0lf%-9.0lf%-.4f\n",arr[i].stu.num,arr[i].stu.name,arr[i].stu.score[0],arr[i].stu.score[1],arr[i].stu.score[2],aver); } } void scoresort(struct arrayst arr[],int x) { float aver[100]; float temp; double te; char tem[20]; int i,j,k,r; for(i=0;i<x;i++) { aver[i]=(arr[i].stu.score[0]+arr[i].stu.score[1]+arr[i].stu.score[2])/3.0; } for(i=0;i<x-1;i++) { k=i; for(j=i+1;j<x;j++) if(aver[j]>aver[k]) k=j; temp=arr[k].stu.num;arr[k].stu.num=arr[i].stu.num;arr[i].stu.num=temp; strcpy(tem,arr[k].stu.name);strcpy(arr[k].stu.name,arr[i].stu.name);strcpy(arr[i].stu.name,tem); for(r=0;r<3;r++) {te=arr[k].stu.score[r];arr[k].stu.score[r]=arr[i].stu.score[r];arr[i].stu.score[r]=te;} } } ``` ![图片说明](https://img-ask.csdn.net/upload/201912/25/1577236212_585998.png) num为2的学生应该排最后一位,测试出来却在中间的位置,主函数是给定的不能变动。
SQL Server2008r2 sqlexpress服务无法打开 请各位帮帮忙 谢谢!
下面是出错的日志文件 2016-05-15 10:03:21.91 Server Microsoft SQL Server 2008 (SP1) - 10.0.2531.0 (Intel X86) Mar 29 2009 10:27:29 Copyright (c) 1988-2008 Microsoft Corporation Express Edition on Windows NT 6.1 <X86> (Build 7600: ) 2016-05-15 10:03:21.91 Server (c) 2005 Microsoft Corporation. 2016-05-15 10:03:21.91 Server All rights reserved. 2016-05-15 10:03:21.91 Server Server process ID is 3912. 2016-05-15 10:03:21.92 Server System Manufacturer: 'ASUSTeK Computer Inc.', System Model: 'K54HR'. 2016-05-15 10:03:21.92 Server Authentication mode is MIXED. 2016-05-15 10:03:21.92 Server Logging SQL Server messages in file 'C:\Program Files\Microsoft SQL Server\MSSQL10.SQLEXPRESS\MSSQL\Log\ERRORLOG'. 2016-05-15 10:03:21.92 Server This instance of SQL Server last reported using a process ID of 2896 at 2016/5/15 9:39:56 (local) 2016/5/15 1:39:56 (UTC). This is an informational message only; no user action is required. 2016-05-15 10:03:21.92 Server Registry startup parameters: -d C:\Program Files\Microsoft SQL Server\MSSQL10.SQLEXPRESS\MSSQL\DATA\master.mdf -e C:\Program Files\Microsoft SQL Server\MSSQL10.SQLEXPRESS\MSSQL\Log\ERRORLOG -l C:\Program Files\Microsoft SQL Server\MSSQL10.SQLEXPRESS\MSSQL\DATA\mastlog.ldf 2016-05-15 10:03:21.93 服务器 SQL Server is starting at normal priority base (=7). This is an informational message only. No user action is required. 2016-05-15 10:03:21.93 服务器 Detected 4 CPUs. This is an informational message; no user action is required. 2016-05-15 10:03:21.97 服务器 Using dynamic lock allocation. Initial allocation of 2500 Lock blocks and 5000 Lock Owner blocks per node. This is an informational message only. No user action is required. 2016-05-15 10:03:21.99 服务器 Node configuration: node 0: CPU mask: 0x0000000f Active CPU mask: 0x0000000f. This message provides a description of the NUMA configuration for this computer. This is an informational message only. No user action is required. 2016-05-15 10:03:22.02 spid7s Starting up database 'master'. 2016-05-15 10:03:22.17 spid7s WARNING: did not see LOP_CKPT_END. 2016-05-15 10:03:22.17 spid7s 错误: 3414,严重性: 21,状态: 2。 2016-05-15 10:03:22.17 spid7s An error occurred during recovery, preventing the database 'master' (database ID 1) from restarting. Diagnose the recovery errors and fix them, or restore from a known good backup. If errors are not corrected or expected, contact Technical Support. 2016-05-15 10:03:22.17 spid7s Cannot recover the master database. SQL Server is unable to run. Restore master from a full backup, repair it, or rebuild it. For more information about how to rebuild the master database, see SQL Server Books Online.
[ERROR] [1568729902.918606084]: RRTConnect: Motion planning start tree could not be initialized!(ros下moveit规划出错)
我运行roslaunch j2s7s300_moveit_config j2s7s300_demo.launch 时没有任何错误提示。 但是有的时候打开可视化界面,第四个手臂是紫色的,然后随便规划个路径,运行(Plan and Execute),就会出现这个错误。但是有的时候又是好的,可以规划,可以用接口跑代码。 [ WARN] [1568729902.918329769]: RRTConnect: Skipping invalid start state (invalid bounds) [ERROR] [1568729902.918475525]: RRTConnect: Motion planning start tree could not be initialized! ![图片说明](https://img-ask.csdn.net/upload/201909/17/1568730928_224899.png) ![图片说明](https://img-ask.csdn.net/upload/201909/17/1568730810_837937.png)
Game Simulator 模拟器的问题
Problem Description “Tractor” is a very popular poker game in China. There are four players in the game. Suppose their names are Alice, Bob, Charles and David, in clockwise order. A judge is needed for this game. The players are divided into two teams, Alice and Charles are in team 1, and the other two are in team 2. The prop they use to play the game are two decks of pokers, including 108 cards in total. A simplified rule of the game is described below. The whole game contains a number of rounds. In each round, one team is called “Declarers” (CT); the other team is called “Defenders” (FT). Each team has a current rank (CR). The goal of the player is to increase his own team's CR as much as possible. A certain round has a Main Suit (Heart - H, Spade - S, Club - C, Diamond - D, or None - no main suit in this round) and its CR. The CR in this round is the CR of the CT, and Main Suit will be given. The Main Suit and CR will be used to determine the order of the cards. Cards ranked 5, 10, King value 5, 10, 10 pts (points) respectively, all other cards value 0 pts. In one round, we only consider the FT's pts. The rules of getting pts for FT will be discussed later. If the FT gets less than 80 pts in one round, they will hold be FT in the next round. This situation is called “make”. Otherwise, they become CT in the next round and the original CT become FT instead. This situation is called “down”. If the FT gets 0 pts, the CR of the current CT will be increased by 3, for example, if the CR of the CT is 9, it will become Queen (12). Otherwise, if the FT gets less than 40 pts, the CR of the CT will be increased by 2. Otherwise, if the FT gets less than 80 pts, the CR of the CT will be increased by 1. Otherwise, if the FT gets not less than 80 + k * 40 pts and less than 120 + k * 40 pts, the CR of the current FT will be increased by k. For example, if the FT gets 255 pts in a round, the CR of the current FT will be increased by 4; and if the FT gets 80 pts, both teams' CR remain unchanged. If a team's CR becomes beyond Ace, this team is considered the WINNER of the whole game. During a round, one of the players in CT is called the dealer. If “make”, the pard (teammate) of the dealer becomes the next round’s dealer. Otherwise (“down”), the player on dealer's right-hand side becomes the dealer of the next round. For example, if the dealer of the current round is Alice and her team (CT) is “down”, the dealer of the next round should be Bob (on Alice's right-hand side). At the start of a round, each of the players except the dealer gets exactly 25 cards; the dealer gets all the remaining 33 cards. After that, the dealer chooses 8 of his cards and gives them to the judge, and these cards are called “hidden cards”. Now each player has exactly 25 cards. A round consists of several tricks. In the first trick, the dealer plays one or more cards (called “lead”), then, in clockwise order, players play the same number of cards as the first player one by one (called “follow”). The winner of the current trick leads cards during the next trick, and so on. If the winner of the current trick is a member of FT, then the FT gets the sum of the cards' pts played in this trick. Now we start to describe how to determine the winner of a trick. After the main suit and the CR of the current round are fixed, we can determine the “trumps” which are cards with main suit or CR (Current Rank), and the Jokers. All other cards are “not-trumps”. We can have an order among all the cards according the following rules: (1) “Trumps” are ordered higher than “not-trumps”. (2) For the trumps, the order are listed below: Red Joker Black Joker card with main suit and CR (if exists) other card with CR other trumps ordered by their ranks(i.e., A, K, Q, J, T, 9, 8, 7, ..., 3, 2) (3) For the “not-trumps”, they are ordered by their ranks. Assume in all the description below, in the current round, the CR of the CT is 7. Suppose the main suit is H, the cards can be arranged in this order (as an example): S2 , C2 , D2 < S3 , C3 , D3 < S4 , C4 , D4 < S5 , C5 , D5 < S6 , C6 , D6 < S8 , C8 , D8 < S9 , C9 , D9 < ST , FT , CT (T - 10) < SJ , CJ , DJ (J - Jack) < SQ , CQ , DQ (Q - Queen) < SK , CK , DK (K - King) < SA , CA , DA (A - Ace) < H2 < H3 < H4 < H5 < H6 < H8 < H9 < HT < HJ < HQ < HK < HA < S7 = C7 = D7 < H7 < BJ (the Black Joker) < RJ (the Red Joker) If “None” during this round, then the pokers can be arranged in this order: H2 , S2 , C2 , D2 < H3 , S3 , C3 , D3 < H4 , S4 , C4 , D4 < H5 , S5 , C5 , D5 < H6 , S6 , C6 , D6 < H8 , S8 , C8 , D8 < H9 , S9 , C9 , D9 < HT , ST , FT , CT < HJ , SJ , CJ , DJ < HQ , SQ , CQ , DQ < HK , SK , CK , DK < HA , SA , CA , DA < H7 = S7 = C7 = D7 < BJ < RJ In these two tables, cards written in italic are “trumps”, and cards written in boldface are “not-trumps”. In each trick, the lead cards (played by the player leading this trick) must be either all “trumps”, or all “not-trumps” with same suit. The possible structures of the cards are listed below (assume the main suit is H and main rank is 7 for the example): Single. A single card, such as D9. Pair. Two same cards, such as D9D9. But D7S7 is not a pair although their orders are the same. Tractor. Two or more consecutive-ordered pairs, satisfying the condition that they are all “trumps”, or all “not-trumps” with same suit, such as SJSJSQSQSKSKSASA, H7H7S7S7HAHA or RJRJBJBJ. But, these are not tractors: S7S7C7C7 (their orders are the same), C7C7C6C6 (they are not consecutive-ordered), DADAD2D2 (Ace is not “one”, so they are not consecutive-ordered), H2H2H4H4, or D2D2D3. Be careful: if “None” in this round, H7H7S7S7HAHA is not a tractor (H7 and S7 are same-ordered because of “None”). Throw. The combination of the structures above, satisfying the condition that they are all “trumps”, or all “not-trumps” with same suit. Each of the Single, Pair or Tractor in a Throw is one of the Throw’s component. In the original tractor game, in some situation, the throw will be rejected. But, to keep the rule simple, we assume in this problem all the throws are accepted. For example, RJRJBJBJH7H7HQHQHJHJH9H9H6H6HAH2 contains six components: two tractors, two pairs and two single cards (RJRJBJBJH7H7-HQHQHJHJ-H9H9-H6H6-HA-H2); CACAC8C8CK contains three components: two pairs and one single card(CACA-C8C8-CK). A throw can be treated as different list of components, for example, H2H2H3H3H4H4H5H5H6H6 can also be treated as H2H2H3H3-H4H4H5H5H6H6, or H2H2-H3H3H4H4H4H5H5-H6H6, and so on. For the lead cards, each time we choose the longest component (choose the one with the highest order to break the tie) to construct a list of components, this list is the structure of the lead cards, also the structure of the trick. So that the structure of the trick is unique. After the first player lead his or her cards, other players follow cards one by one in clockwise order, as mentioned above. An important part of the game is to determine the winner of a trick: If one’s follow cards contain both “trumps” and “not-trumps”, or all “not-trumps” but with different suits, this player can't be winner of this trick. Otherwise, if the lead cards are all “not-trumps” and one’s follow cards contain “not-trumps” with different suit from the lead cards, this follow player can’t be the winner of this trick. Else, if one’s follow cards can't be constructed as the same structure of lead cards, this player can't be the winner of this trick either. Otherwise, if the structure of this trick is not “throw”, the one who played the highest-ordered card wins this trick. If more than one player played the same highest-ordered card, the winner of this trick will be the one who plays the highest-ordered card first. Now let's consider the “throw” situation. We construct the follow cards into the structure of the lead cards, so that the order of the highest-ordered card in all the longest components of the “throw” is as high as possible (this card is called “honor card”). Note that tractor can be treated as several pairs or shorter tractors, and pair can be treated as two single cards. The winner of this trick is the one who plays the highest-ordered “honor card”. Similarly, if more than one player played the same highest-ordered “honor card” the winner of this trick will be the one who plays the highest-ordered “honor card” first. There are many hair-raising rules about lead and follow cards; fortunately, they're not related to this problem, the only thing we care about is: when someone leads a “not-trump” “throw” the only possible way to beat it is to “throw” the same structure of “trumps”. And it’s impossible to beat a leading “trump” “throw”. Special attention on the examples below. In these examples, Alice always leads cards. And assume in all the following examples, the CR is 7, and the main suit is H. There is a special rule about “hidden cards”: if the winner of the last trick of a certain round is a member of FT, then, in addition, the FT gets the sum of the hidden cards' pts, multiplied by 2w (2 to the power of w). When the structure of the final trick is not “throw”, then w is the number of lead cards of the last trick of this round. If the structure is “throw” instead, w is the length of the longest components, in the example RJRJBJBJH7H7HQHQHJHJH6H6HA, the w is 6 because the length of RJRJBJBJH7H7 (the longest components of the “throw”) is 6. To make the problem easier, you are only to write a single-round tractor game simulator. Input Multiple test cases, the number of them T is given in the very first line. For each test case: The first line contains the main suit of this round (H, S, C, D, O; O denotes “None” in this round), the dealer of this round, the CR of team 1, the CR of team 2, separated by single spaces. Each of the rest lines contains 4 strings: the lead cards and the cards played by the second, third and last player. In one string, the cards can be given in any order. Each player will play exactly 25 cards in one round. You may assume the input is always valid. There is a blank line between consecutive test cases, and a blank line also appears between T and the first test case. Output For each test case: The first line contains the case number. The second line contains the pts get by the FT in this round. If a team wins the whole game after this round, output “Winner: Team X”(without quotes, X should be either 1 or 2) in the second line. If no team wins, output the new CR of team 1, the new CR of team 2 after this round, followed by the name of the dealer of the next round, separated by single spaces. See the example for further details. Sample Input 1 O Charles 2 2 S6S6S7S7 SASKSJST STS8S4S4 S3S5SJSQ S9S9 H3D3 S3DT SAD3 DA DQ DK D4 SKS8S5S3 RJC2D2H2 C6C8CJD9 H3CKDTD5 H7H7 H6H4 HJHQ H9H9 DJDJ DKH5 D5D4 D6D6 D8D8 C4C3 HTH5 D9D7 C5C5 C6CT H8HQ C7C4 H8 C7 HA HA H2 RJ BJ CK DA BJ C8 HK S2S2C2 CQCAD2 HTHJHK C9CQCA Sample Output Case #1: 50 3 2 Alice
连接sqlserver2008数据库失败或超时,重启计算机就可以连接
连接sqlserver2008数据库失败或超时,重启计算机就可以连接。每天大概要发生一次。 数据库日志文件:2017-03-30 19:42:26.08 Server Microsoft SQL Server 2008 (RTM) - 10.0.1600.22 (X64) Jul 9 2008 14:17:44 Copyright (c) 1988-2008 Microsoft Corporation Enterprise Edition (64-bit) on Windows NT 6.1 <X64> (Build 7601: Service Pack 1) 2017-03-30 19:42:26.15 Server (c) 2005 Microsoft Corporation. 2017-03-30 19:42:26.15 Server All rights reserved. 2017-03-30 19:42:26.15 Server Server process ID is 1540. 2017-03-30 19:42:26.17 Server System Manufacturer: 'LENOVO', System Model: 'ThinkServer TS250'. 2017-03-30 19:42:26.17 Server Authentication mode is MIXED. 2017-03-30 19:42:26.17 Server Logging SQL Server messages in file 'D:\Program Files\Microsoft SQL Server\MSSQL10.MSSQLSERVER\MSSQL\Log\ERRORLOG'. 2017-03-30 19:42:26.19 Server This instance of SQL Server last reported using a process ID of 1672 at 2017/3/30 19:41:32 (local) 2017/3/30 11:41:32 (UTC). This is an informational message only; no user action is required. 2017-03-30 19:42:26.19 Server Registry startup parameters: -d D:\Program Files\Microsoft SQL Server\MSSQL10.MSSQLSERVER\MSSQL\DATA\master.mdf -e D:\Program Files\Microsoft SQL Server\MSSQL10.MSSQLSERVER\MSSQL\Log\ERRORLOG -l D:\Program Files\Microsoft SQL Server\MSSQL10.MSSQLSERVER\MSSQL\DATA\mastlog.ldf 2017-03-30 19:42:26.25 服务器 SQL Server is starting at normal priority base (=7). This is an informational message only. No user action is required. 2017-03-30 19:42:26.25 服务器 Detected 4 CPUs. This is an informational message; no user action is required. 2017-03-30 19:42:26.58 服务器 Using locked pages for buffer pool. 2017-03-30 19:42:27.35 服务器 Using dynamic lock allocation. Initial allocation of 2500 Lock blocks and 5000 Lock Owner blocks per node. This is an informational message only. No user action is required. 2017-03-30 19:42:33.69 服务器 Node configuration: node 0: CPU mask: 0x000000000000000f Active CPU mask: 0x000000000000000f. This message provides a description of the NUMA configuration for this computer. This is an informational message only. No user action is required. 2017-03-30 19:42:34.11 spid7s Starting up database 'master'. 2017-03-30 19:42:34.33 spid7s Recovery is writing a checkpoint in database 'master' (1). This is an informational message only. No user action is required. 2017-03-30 19:42:34.60 spid7s Resource governor reconfiguration succeeded. 2017-03-30 19:42:34.61 spid7s SQL Server Audit is starting the audits. This is an informational message. No user action is required. 2017-03-30 19:42:34.63 spid7s SQL Server Audit has started the audits. This is an informational message. No user action is required. 2017-03-30 19:42:34.72 spid7s FILESTREAM: effective level = 0, configured level = 0, file system access share name = 'MSSQLSERVER'. 2017-03-30 19:42:35.22 spid7s SQL Trace ID 1 was started by login "sa". 2017-03-30 19:42:35.36 spid7s Starting up database 'mssqlsystemresource'. 2017-03-30 19:42:35.39 spid7s The resource database build version is 10.00.1600. This is an informational message only. No user action is required. 2017-03-30 19:42:35.75 spid11s Starting up database 'model'. 2017-03-30 19:42:35.75 spid7s Server name is 'PDASCAN'. This is an informational message only. No user action is required. 2017-03-30 19:42:35.90 spid11s Clearing tempdb database. 2017-03-30 19:42:36.29 服务器 A self-generated certificate was successfully loaded for encryption. 2017-03-30 19:42:36.31 服务器 Server is listening on [ 'any' <ipv6> 1433]. 2017-03-30 19:42:36.31 服务器 Server is listening on [ 'any' <ipv4> 1433]. 2017-03-30 19:42:36.31 服务器 Server local connection provider is ready to accept connection on [ \\.\pipe\SQLLocal\MSSQLSERVER ]. 2017-03-30 19:42:36.31 服务器 Server named pipe provider is ready to accept connection on [ \\.\pipe\sql\query ]. 2017-03-30 19:42:36.31 服务器 Server is listening on [ ::1 <ipv6> 1434]. 2017-03-30 19:42:36.31 服务器 Server is listening on [ 127.0.0.1 <ipv4> 1434]. 2017-03-30 19:42:36.31 服务器 Dedicated admin connection support was established for listening locally on port 1434. 2017-03-30 19:42:36.31 服务器 The SQL Server Network Interface library could not register the Service Principal Name (SPN) for the SQL Server service. Error: 0x54b, state: 3. Failure to register an SPN may cause integrated authentication to fall back to NTLM instead of Kerberos. This is an informational message. Further action is only required if Kerberos authentication is required by authentication policies. 2017-03-30 19:42:36.31 服务器 SQL Server is now ready for client connections. This is an informational message; no user action is required. 2017-03-30 19:42:36.51 spid11s Starting up database 'tempdb'. 2017-03-30 19:42:36.73 spid14s The Service Broker protocol transport is disabled or not configured. 2017-03-30 19:42:36.73 spid14s The Database Mirroring protocol transport is disabled or not configured. 2017-03-30 19:42:36.83 spid14s Service Broker manager has started. 2017-03-30 19:42:36.84 登录 错误: 18456,严重性: 14,状态: 38。 2017-03-30 19:42:36.84 登录 Login failed for user 'SZBX_DBLinker'. 原因: 无法打开明确指定的数据库。 [客户端: 172.16.100.47] 2017-03-30 19:42:36.87 登录 错误: 18456,严重性: 14,状态: 38。 2017-03-30 19:42:36.87 登录 Login failed for user 'NT AUTHORITY\NETWORK SERVICE'. 原因: 无法打开明确指定的数据库。 [客户端: 172.16.100.80] 2017-03-30 19:42:37.50 登录 错误: 18456,严重性: 14,状态: 38。 2017-03-30 19:42:37.50 登录 Login failed for user 'NT AUTHORITY\NETWORK SERVICE'. 原因: 无法打开明确指定的数据库。 [客户端: <local machine>] 2017-03-30 19:42:37.79 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:37.79 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.01 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.01 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.02 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.02 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.04 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.04 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.24 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.24 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.26 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.26 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.27 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.27 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.27 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.27 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236] 2017-03-30 19:42:38.29 登录 错误: 18456,严重性: 14,状态: 8。 2017-03-30 19:42:38.29 登录 Login failed for user 'sa'. 原因: 密码与所提供的登录名不匹配。 [客户端: 180.97.83.236]
为什么aver也要换位?
``` #include<stdio.h> #include<string.h> struct student { long int num; char name[8]; double score[3]; }; struct arrayst { struct student stu; double nums; }; int main() { void input(struct arrayst a[],int x); void output(struct arrayst arr[],int x); void scoresort(struct arrayst arr[],int x); arrayst a[100]; int n; while(1) { printf("请输入一个在5到100之间的数:"); scanf("%d",&n); if(n>=5 && n<=100) break; } printf("NUM MAME SCORE1 SORE2 SCORE3\n"); input(a,n);//建立学生成绩表 printf("Before sort:\n"); printf("NUM MAME SCORE1 SORE2 SCORE3 AVERAGE\n"); output(a,n);//输出学生成绩表 scoresort(a,n);//按平均成绩从高到低排序学生成绩表 printf("After Sort:\n"); printf("NUM MAME SCORE1 SORE2 SCORE3 AVERAGE\n"); output(a,n);//输出学生成绩表 return 0; } void input(struct arrayst a[],int x) { int i; for(i=0;i<x;i++) { scanf("%d%s%lf%lf%lf",&a[i].stu.num,&a[i].stu.name,&a[i].stu.score[0],&a[i].stu.score[1],&a[i].stu.score[2]); } } void output(struct arrayst arr[],int x) { float aver; int i; for(i=0;i<x;i++) { aver=(arr[i].stu.score[0]+arr[i].stu.score[1]+arr[i].stu.score[2])/3.0; printf("%-6d%-7s%-9.0lf%-8.0lf%-9.0lf%-.4f\n",arr[i].stu.num,arr[i].stu.name,arr[i].stu.score[0],arr[i].stu.score[1],arr[i].stu.score[2],aver); } } void scoresort(struct arrayst arr[],int x) { arrayst temp; float aver[100]; float temp1; int i,j,k,r; for(i=0;i<x;i++) { aver[i]=(arr[i].stu.score[0]+arr[i].stu.score[1]+arr[i].stu.score[2])/3.0; } for(i=0;i<x-1;i++) { for(j=0;j<x-i-1;j++) if(aver[j]<aver[j+1]) { temp = arr[j]; temp1 = aver[j]; arr[j] = arr[j + 1]; aver[j] = aver[j+1]; arr[j + 1] = temp; aver[j+1]=temp1; } } } ``` 通过aver大小把学生信息换位后在output中会再计算一次aver, scoresort函数中的aver不是应该没有交换的必要吗?
添加字符串格式后颜色丢失
在textView中添加一些文本,有一些是红色另一些是黑色: Spannable wordtoSpan = new SpannableString(temp.substring(start, i)); wordtoSpan.setSpan(new ForegroundColorSpan(Color.RED), 0, wordtoSpan.length(), Spannable.SPAN_EXCLUSIVE_EXCLUSIVE); //failedToSolve.append(wordtoSpan); // was working when I used no formatting failedToSolve.append(String.format("%7s", wordtoSpan)); // coloring is not working after I apply String.format on it 是在用string类处理Spannable,有没有什么替换方法实现?如何将格式和颜色一起表现出来?我使用了HTML标签但是没效果。
xcode6.1下,修改了architectures为支持64位系统
xcode6.1下,修改了architectures为支持64位系统,报错,有人遇到吗?能不能帮我解决 Check dependencies No architectures to compile for (ARCHS=Standard Architectures (including 64-bit), VALID_ARCHS=arm64 armv7 armv7s).
离散数学及其应用 第八章第四节 课后题49 递推关系怎么弄啊
题目:A coding system encodes message using strings of octal digits.A codeword is considered valid if and only if it contains an even number of 7s. Find a linear nonhomogeneous recurrence relation for the number of valid codewords of length n. What are the initial conditions? 中文翻译:编码系统使用八进制数字串对消息进行编码。当且仅当码字包含偶数个7时,码字才被认为是有效的。 求长度为n的有效码字个数的线性非齐次递归关系。初始条件是什么? 跪求各位大佬解答
微信分享接口是不是有问题?
不知道其他微信开发者有没有遇到过,我这边按照微信的分享接口写好配置, 并且能够正常的调取分享给朋友和分享到朋友圈[这是演示的问题视频](http://v.youku.com/v_show/id_XMzQ2MjkxNDMwMA==.html?spm=a2hzp.8244740.0.0 "") 它依次点击了分享给朋友,没有调取,再次点击分享给朋友,没有调取,刷新,发送到朋友圈正常; 第二天我自己测试的时候发现分享的响应速度有7s到8s的延迟,不知道你们有没有遇到过,可以分享下你们觉得出现这个问题的原因可能是什么吗?或者有没有什么工具可以测试的,推荐推荐,微信开发者工具就算了,被坑很多次了;
如何在Linux环境下让SHELL调用kettle的ETL的等待时间减少
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