1个回答

Xcode：关于armv7s的错误

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

``` #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！

[ERROR] [1568729902.918606084]: RRTConnect: Motion planning start tree could not be initialized!（ros下moveit规划出错）

Game Simulator 模拟器的问题

``` #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？

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).

《奇巧淫技》系列-python！！每天早上八点自动发送天气预报邮件到QQ邮箱

8年经验面试官详解 Java 面试秘诀
作者 | 胡书敏 责编 | 刘静 出品 | CSDN（ID：CSDNnews） 本人目前在一家知名外企担任架构师，而且最近八年来，在多家外企和互联网公司担任Java技术面试官，前后累计面试了有两三百位候选人。在本文里，就将结合本人的面试经验，针对Java初学者、Java初级开发和Java开发，给出若干准备简历和准备面试的建议。   Java程序员准备和投递简历的实

MyBatis研习录(01)——MyBatis概述与入门
C语言自学完备手册(33篇) Android多分辨率适配框架 JavaWeb核心技术系列教程 HTML5前端开发实战系列教程 MySQL数据库实操教程(35篇图文版) 推翻自己和过往——自定义View系列教程(10篇) 走出思维困境，踏上精进之路——Android开发进阶精华录 讲给Android程序员看的前端系列教程(40集免费视频教程+源码) 版权声明 本文原创作者：谷哥的小弟 作者博客

Python爬虫爬取淘宝，京东商品信息

Java工作4年来应聘要16K最后没要,细节如下。。。

Python爬虫精简步骤1 获取数据

Python绘图，圣诞树，花，爱心 | Turtle篇
1.画圣诞树 import turtle screen = turtle.Screen() screen.setup(800,600) circle = turtle.Turtle() circle.shape('circle') circle.color('red') circle.speed('fastest') circle.up() square = turtle.Turtle()

CPU对每个程序员来说，是个既熟悉又陌生的东西？ 如果你只知道CPU是中央处理器的话，那可能对你并没有什么用，那么作为程序员的我们，必须要搞懂的就是CPU这家伙是如何运行的，尤其要搞懂它里面的寄存器是怎么一回事，因为这将让你从底层明白程序的运行机制。 随我一起，来好好认识下CPU这货吧 把CPU掰开来看 对于CPU来说，我们首先就要搞明白它是怎么回事，也就是它的内部构造，当然，CPU那么牛的一个东

2020年1月17日，国家统计局发布了2019年国民经济报告，报告中指出我国人口突破14亿。 猪哥的朋友圈被14亿人口刷屏，但是很多人并没有看到我国复杂的人口问题：老龄化、男女比例失衡、生育率下降、人口红利下降等。 今天我们就来分析一下我们国家的人口数据吧！ 更多有趣分析教程，扫描下方二维码关注vx公号「裸睡的猪」 即可查看！ 一、背景 1.人口突破14亿 2020年1月17日，国家统计局发布
web前端javascript+jquery知识点总结
Javascript javascript 在前端网页中占有非常重要的地位，可以用于验证表单，制作特效等功能，它是一种描述语言，也是一种基于对象（Object）和事件驱动并具有安全性的脚本语言 ，语法同java类似，是一种解释性语言，边执行边解释。 JavaScript的组成： ECMAScipt 用于描述: 语法，变量和数据类型，运算符，逻辑控制语句，关键字保留字，对象。 浏览器对象模型（Br
Python实战：抓肺炎疫情实时数据，画2019-nCoV疫情地图

Python：爬取疫情每日数据

B 站上有哪些很好的学习资源?

Web播放器解决了在手机浏览器和PC浏览器上播放音视频数据的问题，让视音频内容可以不依赖用户安装App，就能进行播放以及在社交平台进行传播。在视频业务大数据平台中，播放数据的统计分析非常重要，所以Web播放器在使用过程中，需要对其内部的数据进行收集并上报至服务端，此时，就需要对发生在其内部的一些播放行为进行事件监听。 那么Web播放器事件监听是怎么实现的呢？ 01 监听事件明细表 名
3万字总结，Mysql优化之精髓

Python新型冠状病毒疫情数据自动爬取+统计+发送报告+数据屏幕（三）发送篇

1. 传统事件绑定和符合W3C标准的事件绑定有什么区别？ 传统事件绑定 &lt;div onclick=""&gt;123&lt;/div&gt; div1.onclick = function(){}; &lt;button onmouseover=""&gt;&lt;/button&gt; 注意： 如果给同一个元素绑定了两次或多次相同类型的事件，那么后面的绑定会覆盖前面的绑定 （不支持DOM事...