POJ 2406 其实就是一个简单的kmp应用: ans = n % (n - f[n]) == 0 ? n / (n - f[n]) : 1 其中f是失配函数 //#pragma comment(linker, "/STACK:1677721600") #include <map> #include <set> #include <stack> #include <queue> #include <cmath> #inclu…

http://poj.org/problem?id=2406 #include<cstdio> #include<cstring> #include<algorithm> #define maxn 1000010 using namespace std; char s[maxn]; int next[maxn]; void get(char *s,int *next) { int j,k; next[]=-; j=; k=-; int kk=strlen(s); whi…

Period Time Limit: 3000MSMemory Limit: 30000K Total Submissions: 12089Accepted: 5656 Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefi…

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…

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 =…

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…

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…

Power Strings 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 think of concatenation as…

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…

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 = "…