poj1961---KMP

//和poj2406同样处理//一个失败函数就行//因为s=a^n，所以如果i%(i-f[i])==0，那么其a的长度就是i-f[i]了#include <iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 1000005int n,m,t,p,end;char str[N];int f[N];int main(){    end++;    while(scanf("%d",&n),n)    {        scanf("%s",str);        int i,j,k;        f[0]=j=-1,i=0;        while(i<n)        {            if(j==-1||str[i]==str[j])            {                ++i,++j;                f[i]=j;            }            else j=f[j];        }        printf("Test case #%d\n",end++);        for(i=2;i<=n;i++)        {            if(i%(i-f[i])==0&&i/(i-f[i])!=1)            {                printf("%d %d\n",i,i/(i-f[i]));            }        }        puts("");    }    return 0;}