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Queue Sequence 队列的算法
Problem Description There's a queue obeying the first in first out rule. Each time you can either push a number into the queue (+i), or pop a number out from the queue (-i). After a series of operation, you get a sequence (e.g. +1 -1 +2 +4 -2 -4). We call this sequence a queue sequence. Now you are given a queue sequence and asked to perform several operations: 1. insert p First you should find the smallest positive number (e.g. i) that does not appear in the current queue sequence, then you are asked to insert the +i at position p (position starts from 0). For -i, insert it into the right most position that result in a valid queue sequence (i.e. when encountered with element -x, the front of the queue should be exactly x). For example, (+1 -1 +3 +4 -3 -4) would become (+1 +2 -1 +3 +4 -2 -3 -4) after operation 'insert 1'. 2. remove i Remove +i and -i from the sequence. For example, (+1 +2 -1 +3 +4 -2 -3 -4) would become (+1 +2 -1 +4 -2 -4) after operation 'remove 3'. 3. query i Output the sum of elements between +i and -i. For example, the result of query 1, query 2, query 4 in sequence (+1 +2 -1 +4 -2 -4) is 2, 3(obtained by -1 + 4), -2 correspond. Input There are less than 25 test cases. Each case begins with a number indicating the number of operations n (1 ≤ n ≤ 100000). The following n lines with be 'insert p', 'remove i' or 'query i'(0 ≤ p ≤ length (current sequence), 1 ≤ i, i is granted to be in the sequence). In each case, the sequence is empty initially. The input is terminated by EOF. Output Before each case, print a line "Case #d:" indicating the id of the test case. After each operation, output the sum of elements between +i and -i. Sample Input 10 insert 0 insert 1 query 1 query 2 insert 2 query 2 remove 1 remove 2 insert 2 query 3 6 insert 0 insert 0 remove 2 query 1 insert 1 query 2 Sample Output Case #1: 2 -1 2 0 Case #2: 0 -1
Queue Sequence 队列的顺序
Problem Description There's a queue obeying the first in first out rule. Each time you can either push a number into the queue (+i), or pop a number out from the queue (-i). After a series of operation, you get a sequence (e.g. +1 -1 +2 +4 -2 -4). We call this sequence a queue sequence. Now you are given a queue sequence and asked to perform several operations: 1. insert p First you should find the smallest positive number (e.g. i) that does not appear in the current queue sequence, then you are asked to insert the +i at position p (position starts from 0). For -i, insert it into the right most position that result in a valid queue sequence (i.e. when encountered with element -x, the front of the queue should be exactly x). For example, (+1 -1 +3 +4 -3 -4) would become (+1 +2 -1 +3 +4 -2 -3 -4) after operation 'insert 1'. 2. remove i Remove +i and -i from the sequence. For example, (+1 +2 -1 +3 +4 -2 -3 -4) would become (+1 +2 -1 +4 -2 -4) after operation 'remove 3'. 3. query i Output the sum of elements between +i and -i. For example, the result of query 1, query 2, query 4 in sequence (+1 +2 -1 +4 -2 -4) is 2, 3(obtained by -1 + 4), -2 correspond. Input There are less than 25 test cases. Each case begins with a number indicating the number of operations n (1 ≤ n ≤ 100000). The following n lines with be 'insert p', 'remove i' or 'query i'(0 ≤ p ≤ length (current sequence), 1 ≤ i, i is granted to be in the sequence). In each case, the sequence is empty initially. The input is terminated by EOF. Output Before each case, print a line "Case #d:" indicating the id of the test case. After each operation, output the sum of elements between +i and -i. Sample Input 10 insert 0 insert 1 query 1 query 2 insert 2 query 2 remove 1 remove 2 insert 2 query 3 6 insert 0 insert 0 remove 2 query 1 insert 1 query 2 Sample Output Case #1: 2 -1 2 0 Case #2: 0 -1
Queue Sequence 序列的问题
Problem Description There's a queue obeying the first in first out rule. Each time you can either push a number into the queue (+i), or pop a number out from the queue (-i). After a series of operation, you get a sequence (e.g. +1 -1 +2 +4 -2 -4). We call this sequence a queue sequence. Now you are given a queue sequence and asked to perform several operations: 1. insert p First you should find the smallest positive number (e.g. i) that does not appear in the current queue sequence, then you are asked to insert the +i at position p (position starts from 0). For -i, insert it into the right most position that result in a valid queue sequence (i.e. when encountered with element -x, the front of the queue should be exactly x). For example, (+1 -1 +3 +4 -3 -4) would become (+1 +2 -1 +3 +4 -2 -3 -4) after operation 'insert 1'. 2. remove i Remove +i and -i from the sequence. For example, (+1 +2 -1 +3 +4 -2 -3 -4) would become (+1 +2 -1 +4 -2 -4) after operation 'remove 3'. 3. query i Output the sum of elements between +i and -i. For example, the result of query 1, query 2, query 4 in sequence (+1 +2 -1 +4 -2 -4) is 2, 3(obtained by -1 + 4), -2 correspond. Input There are less than 25 test cases. Each case begins with a number indicating the number of operations n (1 ≤ n ≤ 100000). The following n lines with be 'insert p', 'remove i' or 'query i'(0 ≤ p ≤ length (current sequence), 1 ≤ i, i is granted to be in the sequence). In each case, the sequence is empty initially. The input is terminated by EOF. Output Before each case, print a line "Case #d:" indicating the id of the test case. After each operation, output the sum of elements between +i and -i. Sample Input 10 insert 0 insert 1 query 1 query 2 insert 2 query 2 remove 1 remove 2 insert 2 query 3 6 insert 0 insert 0 remove 2 query 1 insert 1 query 2 Sample Output Case #1: 2 -1 2 0 Case #2: 0 -1
queue中存放list的问题

queue<int>q 是不是不能用q[i]=i对队列赋值，只能用q.push() queue<int>q(10） 又和上面的写法有什么区别？ 这种写法可以q[i]=i赋值吧？ 这里的q都是什么含义啊？指针还是什么？ 初学者求解答...
sparkstreaming中的job queue问题？

VS2019中Stack around the variable 'order' was corrupted报错求问
main函数代码如下： int main() { char order[8]; int result[100]={0}; int size , data , node , q=0; priorityQueue queue; cin>>size; for (int k=0 ; k<size ; ++k) { cin>>order; if ( order[0] == 'i' ) { cin>>data; queue.enQueue(data); } else if ( order[0] == 'f' ) { cin>>data; result[q]=queue.find(data); q += 1; } else if ( order[0] == 'd' ) { cin>>node>>data; queue.decrease(node,data); } } for (int k=0 ; result[k]!=0; ++k) cout<<result[k]<<endl; return 0; } 我把开头定义字符数组的部分写成 char *order = new char [8] 就不会出错了。 或者我在开头同时输出order和&q的地址过后也不会出这个错误（我这个是在code::block上写的，输出结果没错，VS上没试过）。 如果直接跑的话VS上会报标题所示”Stack around the variable 'order' was corrupted“的错误，在code::block上表现为q会突然变成0. 是内存溢出的问题？？求教
django中重写logging.Handle的emit，在里面实现日志的保存，但是在配置时一直报错，好像不能配置自定义的handlers

Jumping the Queue 队列解决的办法
Problem Description The beginning of a winter break near Spring Festival is always the beginning of a peak period of transportation. If you have ever tried to get a train ticket at that time, you must have witnessed the endless queues in front of every ticket box window. If a guy has seen his friend in a queue, then it is very much likely that this lucky guy might go straight to his friend and ask for a favor. This is called "jumping the queue". It is unfair to the rest of the people in the line, but, it is life. Your task is to write a program that simulates such a queue with people jumping in every now and then, assume that, if one in the queue has several friends asking for favors, he would arrange their requests in a queue of his own. Input There will contain one or more test cases. Each test case begins with the number of groups n (1<= n <=1000). Then n group descriptions follow, each one consisting of the number of friends belonging to the group and those people's distinct names. A group is a friend group. People in one group are friend with each other. A name is a string of up to 4 characters chosen from {A, B, ..., Z, a, b, ..., z}. A group may consist of up to 1000 friends. You may assume that there is no one belong to two different groups. Finally, a list of commands follows. There are three different kinds of commands: ● ENQUEUE X - Mr. or Ms. X goes into the queue ● DEQUEUE - the first person gets the ticket and leave the queue ● STOP - end of test case The input will be terminated by a value of 0 for n. Output For each test case, first print a line saying "Scenario #k", where k is the number of the test case. Then, for each DEQUEUE command, print the person who just gets a ticket on a single line. Print a blank line between two test cases, but no extra line at the end of output. Sample Input 2 3 Ann Bob Joe 3 Zoe Jim Fat ENQUEUE Ann ENQUEUE Zoe ENQUEUE Bob ENQUEUE Jim ENQUEUE Joe ENQUEUE Fat DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 Anny Jack Jean Bill Jane 6 Eva Mike Ron Sony Geo Zoro ENQUEUE Anny ENQUEUE Eva ENQUEUE Jack ENQUEUE Jean ENQUEUE Bill ENQUEUE Jane DEQUEUE DEQUEUE ENQUEUE Mike ENQUEUE Ron DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0 Sample Output Scenario #1 Ann Bob Joe Zoe Jim Fat Scenario #2 Anny Jack Jean Bill Jane Eva
Jumping the Queue 是怎么写的呢
Problem Description The beginning of a winter break near Spring Festival is always the beginning of a peak period of transportation. If you have ever tried to get a train ticket at that time, you must have witnessed the endless queues in front of every ticket box window. If a guy has seen his friend in a queue, then it is very much likely that this lucky guy might go straight to his friend and ask for a favor. This is called "jumping the queue". It is unfair to the rest of the people in the line, but, it is life. Your task is to write a program that simulates such a queue with people jumping in every now and then, assume that, if one in the queue has several friends asking for favors, he would arrange their requests in a queue of his own. Input There will contain one or more test cases. Each test case begins with the number of groups n (1<= n <=1000). Then n group descriptions follow, each one consisting of the number of friends belonging to the group and those people's distinct names. A group is a friend group. People in one group are friend with each other. A name is a string of up to 4 characters chosen from {A, B, ..., Z, a, b, ..., z}. A group may consist of up to 1000 friends. You may assume that there is no one belong to two different groups. Finally, a list of commands follows. There are three different kinds of commands: ● ENQUEUE X - Mr. or Ms. X goes into the queue ● DEQUEUE - the first person gets the ticket and leave the queue ● STOP - end of test case The input will be terminated by a value of 0 for n. Output For each test case, first print a line saying "Scenario #k", where k is the number of the test case. Then, for each DEQUEUE command, print the person who just gets a ticket on a single line. Print a blank line between two test cases, but no extra line at the end of output. Sample Input 2 3 Ann Bob Joe 3 Zoe Jim Fat ENQUEUE Ann ENQUEUE Zoe ENQUEUE Bob ENQUEUE Jim ENQUEUE Joe ENQUEUE Fat DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 Anny Jack Jean Bill Jane 6 Eva Mike Ron Sony Geo Zoro ENQUEUE Anny ENQUEUE Eva ENQUEUE Jack ENQUEUE Jean ENQUEUE Bill ENQUEUE Jane DEQUEUE DEQUEUE ENQUEUE Mike ENQUEUE Ron DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0 Sample Output Scenario #1 Ann Bob Joe Zoe Jim Fat Scenario #2 Anny Jack Jean Bill Jane Eva

jms采用p2p方式，在web应用中配置了Servlet实现了MessageListener接口，用于获取queue中信息 另写一测试java程序，模拟生产者 运行该web应用和java程序，一切都OK，onMessage方法可以正确接收消息队列中的消息 现在有如下的问题和疑惑： 停止java程序，不再给消息队列中添加数据，然后重启tomcat，Servlet中onMessage方法仍然会把以前消息队列中 的数据又重新显示，我的疑问是，在p2p模式下，queue中的消息何时会被清除，按照我测试的情况发现，并不是有消费者消费 了queue中的消息，消息就会在MOM中清除了，是否要在消费者处理该信息时候进行消息的应答才可以删除queue中的信息呢，本人初学jms，请大虾们指点，谢谢 ps:有些文章介绍队列保留着消息，直到它们被消费，这我和的测试结果不一样
C++中queue能存数组吗？

C++的STL中关于priority_queue算子的详细定义
priority queue的模板参数为<元素类型，容器类型，比较算子> 其中默认的算子为less，我在网上看到的less的解释是：小的元素往前排，大的往后排，出队时序列尾的元素出队 我就是不太懂这里为什么要是对尾的元素先出队，也就是q.top()返回的是队尾元素，且q.pop()删除的也是队尾元素——这不正好跟队列的定义完全相反了么？？
Queue Sequence
roblem Description There's a queue obeying the first in first out rule. Each time you can either push a number into the queue (+i), or pop a number out from the queue (-i). After a series of operation, you get a sequence (e.g. +1 -1 +2 +4 -2 -4). We call this sequence a queue sequence. Now you are given a queue sequence and asked to perform several operations: 1. insert p First you should find the smallest positive number (e.g. i) that does not appear in the current queue sequence, then you are asked to insert the +i at position p (position starts from 0). For -i, insert it into the right most position that result in a valid queue sequence (i.e. when encountered with element -x, the front of the queue should be exactly x). For example, (+1 -1 +3 +4 -3 -4) would become (+1 +2 -1 +3 +4 -2 -3 -4) after operation 'insert 1'. 2. remove i Remove +i and -i from the sequence. For example, (+1 +2 -1 +3 +4 -2 -3 -4) would become (+1 +2 -1 +4 -2 -4) after operation 'remove 3'. 3. query i Output the sum of elements between +i and -i. For example, the result of query 1, query 2, query 4 in sequence (+1 +2 -1 +4 -2 -4) is 2, 3(obtained by -1 + 4), -2 correspond. Input There are less than 25 test cases. Each case begins with a number indicating the number of operations n (1 ≤ n ≤ 100000). The following n lines with be 'insert p', 'remove i' or 'query i'(0 ≤ p ≤ length (current sequence), 1 ≤ i, i is granted to be in the sequence). In each case, the sequence is empty initially. The input is terminated by EOF. Output Before each case, print a line "Case #d:" indicating the id of the test case. After each operation, output the sum of elements between +i and -i. Sample Input 10 insert 0 insert 1 query 1 query 2 insert 2 query 2 remove 1 remove 2 insert 2 query 3 6 insert 0 insert 0 remove 2 query 1 insert 1 query 2 Sample Output Case #1: 2 -1 2 0 Case #2: 0 -1
c++库中 的 priority_queue 构建 小根堆
c++库中 的 priority_queue 构建 小根堆 后 ，top函数 返回的是 const对象。。。怎么变成非const对象 无法从“const HuffmanNode<char>”转换为“HuffmanNode<char>”
Jumping the Queue 怎么弄呢
Problem Description The beginning of a winter break near Spring Festival is always the beginning of a peak period of transportation. If you have ever tried to get a train ticket at that time, you must have witnessed the endless queues in front of every ticket box window. If a guy has seen his friend in a queue, then it is very much likely that this lucky guy might go straight to his friend and ask for a favor. This is called "jumping the queue". It is unfair to the rest of the people in the line, but, it is life. Your task is to write a program that simulates such a queue with people jumping in every now and then, assume that, if one in the queue has several friends asking for favors, he would arrange their requests in a queue of his own. Input There will contain one or more test cases. Each test case begins with the number of groups n (1<= n <=1000). Then n group descriptions follow, each one consisting of the number of friends belonging to the group and those people's distinct names. A group is a friend group. People in one group are friend with each other. A name is a string of up to 4 characters chosen from {A, B, ..., Z, a, b, ..., z}. A group may consist of up to 1000 friends. You may assume that there is no one belong to two different groups. Finally, a list of commands follows. There are three different kinds of commands: ● ENQUEUE X - Mr. or Ms. X goes into the queue ● DEQUEUE - the first person gets the ticket and leave the queue ● STOP - end of test case The input will be terminated by a value of 0 for n. Output For each test case, first print a line saying "Scenario #k", where k is the number of the test case. Then, for each DEQUEUE command, print the person who just gets a ticket on a single line. Print a blank line between two test cases, but no extra line at the end of output. Sample Input 2 3 Ann Bob Joe 3 Zoe Jim Fat ENQUEUE Ann ENQUEUE Zoe ENQUEUE Bob ENQUEUE Jim ENQUEUE Joe ENQUEUE Fat DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 Anny Jack Jean Bill Jane 6 Eva Mike Ron Sony Geo Zoro ENQUEUE Anny ENQUEUE Eva ENQUEUE Jack ENQUEUE Jean ENQUEUE Bill ENQUEUE Jane DEQUEUE DEQUEUE ENQUEUE Mike ENQUEUE Ron DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0 Sample Output Scenario #1 Ann Bob Joe Zoe Jim Fat Scenario #2 Anny Jack Jean Bill Jane Eva
Team Queue
Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for example. In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue. Your task is to write a program that simulates such a team queue. Input The input will contain one or more test cases. Each test case begins with the number of teams t (1<=t<=1000). Then t team descriptions follow, each one consisting of the number of elements belonging to the team and the elements themselves. Elements are integers in the range 0 - 999999. A team may consist of up to 1000 elements. Finally, a list of commands follows. There are three different kinds of commands: ENQUEUE x - enter element x into the team queue DEQUEUE - process the first element and remove it from the queue STOP - end of test case The input will be terminated by a value of 0 for t. Output For each test case, first print a line saying "Scenario #k", where k is the number of the test case. Then, for each DEQUEUE command, print the element which is dequeued on a single line. Print a blank line after each test case, even after the last one. Sample Input 2 3 101 102 103 3 201 202 203 ENQUEUE 101 ENQUEUE 201 ENQUEUE 102 ENQUEUE 202 ENQUEUE 103 ENQUEUE 203 DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 259001 259002 259003 259004 259005 6 260001 260002 260003 260004 260005 260006 ENQUEUE 259001 ENQUEUE 260001 ENQUEUE 259002 ENQUEUE 259003 ENQUEUE 259004 ENQUEUE 259005 DEQUEUE DEQUEUE ENQUEUE 260002 ENQUEUE 260003 DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0 Sample Output Scenario #1 101 102 103 201 202 203 Scenario #2 259001 259002 259003 259004 259005 260001
Printer Queue

Java学习的正确打开方式

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