LA 3026 && POJ 1961 Period（KMP求前缀的最短循环节）

题意：

给定一个长度为n 的字符串s，求它的每个前缀的最短循环节？

思路：

我们想一下KMP的next 函数。

next[i]表示 由S[0],S[1],,,,S[i] 构成的字符串的最大前缀长度 使得前缀等于后缀。

那么这个问题就很好办了。

假设当前枚举到i 位置了。（目前字符串的长度是i）

假设最短循环节是X。

那么next[i] 一定是i - X；

所以我们只需要判断i 是否是i- Next[i]倍数即可。

 注意 next[i]必须大于0

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 1000000 + 10;

char s[maxn];
int n, ks;
int Next[maxn];

void get_Next(){
Next[0] = Next[1] = 0;
int j = 0;
for (int i = 1; i < n; ++i){
j = Next[i];
while(j > 0 && s[i] != s[j]) j = Next[j];
if (s[i] == s[j]) Next[i+1] = j+1;
else Next[i+1] = 0;
}
}
int main(){

while(~scanf("%d",&n) && n){

scanf("%s",s);
get_Next();
printf("Test case #%d\n",++ks);
for (int i = 2; i <= n; ++i){
if (!Next[i])continue;
if (i % (i-Next[i]) == 0){
printf("%d %d\n",i, i/ (i-Next[i]));
}

}
putchar('\n');
}

return 0;
}

Period Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 17547   Accepted: 8460 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 prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK ,that is A concatenated K times, for some string A. Of course, we also want to know the period K. Input The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the  number zero on it. Output For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case. Sample Input 3 aaa 12 aabaabaabaab 0 Sample Output Test case #1 2 2 3 3 Test case #2 2 2 6 2 9 3 12 4 Source Southeastern Europe 2004

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