题意:给一个字符串S长度不超过10^6,求最大的n使得S由n个相同的字符串a连接而成,如:"ababab"则由n=3个"ab"连接而成,"aaaa"由n=4个"a"连接而成,"abcd"则由n=1个"abcd"连接而成. 利用KMP算法,求字符串的特征向量next,若len可以被len - next[len]整除,则最大循环次数n为len/(len - next[len]),否则为1.…

Power Strings Time Limit: 3000MSMemory Limit: 65536K Total Submissions: 29663Accepted: 12387 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef…

http://poj.org/problem?id=2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27003   Accepted: 11311 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = &q…

Power Strings   Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 47748   Accepted: 19902 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = &quo…

点击打开链接 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27368   Accepted: 11454 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b =…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28102   Accepted: 11755 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

                                                                                                  Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 38038   Accepted: 15740 Description Given two strings a and b we define a*b t…

题目链接:http://poj.org/problem?id=2406 Time Limit: 3000MS Memory Limit: 65536K Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we thi…

传送门 http://poj.org/problem?id=2406 题目就是求循环了几次. 记得如果每循环输出为1.... #include<cstdio> #include<cstring> const int MAXN=1000000+10; char P[MAXN]; int f[MAXN]; int n,m; void getFail() { int i,j; f[0]=f[1]=0; for(i=1;i<n;i++) { j=f[i]; while(j &…

本题是计算一个字符串能完整分成多少一模一样的子字符串. 原来是使用KMP的next数组计算出来的,一直都认为是能够利用next数组的.可是自己想了非常久没能这么简洁地总结出来,也仅仅能查查他人代码才恍然大悟,原来能够这么简单地区求一个周期字符串的最小周期的. 有某些大牛建议说不应该參考代码或者解题报告,可是这些大牛却没有给出更加有效的学习方法,比方不懂KMP.难倒不应该去看?要自己想出KMP来吗?我看不太可能有哪位大牛能够直接自己"又一次创造出KMP"来吧. 好吧.不说"创造…