Problem Description Chen, Adrian (November 7, 2013). "Doge Is An Ac- tually Good Internet Meme. Wow.". Gawker. Retrieved November 22, 2013. Doge is an Internet meme that became popular in 2013. The meme typically con- sists of a picture of a Shi…

Wow! Such Doge! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2351    Accepted Submission(s): 1445 Problem Description Chen, Adrian (November 7, 2013). “Doge Is An Ac- tually Good Internet Mem…

Problem Description I believe many people are the fans of prison break. How clever Michael is!! In order that the message won't be found by FBI easily, he usually send code letters to Sara by a paper crane. Hence, the paper crane is Michael in the he…

HDU 4847 Wow! Such Doge! pid=4847" style="">题目链接 题意:给定文本,求有几个doge,不区分大写和小写 思路:水题.直接一个个读字符每次推断就可以 代码: #include <stdio.h> #include <string.h> char c; char a[5]; int main() { a[5] = '\0'; int ans = 0; while ((c = getchar()) != E…

http://acm.hdu.edu.cn/showproblem.php?pid=4847 Wow! Such Doge! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4847 Description Chen, Adrian (November 7, 2013). “Doge Is An Ac- tually Good Inter…

Wow! Such Doge! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3816    Accepted Submission(s): 2254 Problem Description Chen, Adrian (November 7, 2013). “Doge Is An Ac- tually Good Internet Mem…

HDOJ(HDU).1035 Robot Motion [从零开始DFS(4)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (DFS)…

HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开始DFS…

HDU 4850 Wow! Such String! 题目链接 题意:求50W内的字符串.要求长度大于等于4的子串,仅仅出现一次 思路:须要推理.考虑4个字母的字符串,一共同拥有26^4种,这些由这些字符串.假设一个字符串末尾加上一个字符.能够变成还有一个字符串的话,就当作这有一条边,每多一个字符多一个结点,那么对于这道题目,一共就能有26^4 + 3条边,在加上尾巴能够多放3个,一共是26^4+3个边.这些边所有连起来就是要的字符串,这样就能够知道每一个节点会经过的次数为26,这样就仅仅要考虑…

作者:jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4080264.html 题目链接:hdu 4850 Wow! Such String! 欧拉回路 长度为4的由26个字母组成的字符串一共有$4^{26}$种,从aaaa开始,在加上结尾的aaa那么该字符串长度为$4^{26}+3$.当字符串i的后三个字母和字符串j的前三个字母相同则ij有一条边,遍历所有的边可以构成一个欧拉回路. 首先构造aaaabbbb...zzzz的字符串,然后依次向结尾…