Uvalive

参考：http://www.cnblogs.com/jackge/archive/2013/01/05/2846006.html

总结一下，如果对于next数组中的 i， 符合 i % ( i - next[i] ) == 0 && next[i] != 0 , 则说明字符串循环，而且

循环节长度为:   i - next[i]

循环次数为:       i / ( i - next[i] )

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#pragma comment(linker, "/STACK:102400000,102400000")
#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("b.txt", "w", stdout);
#define maxn 1000000000
#define N 2510
#define mod 1000000000
using namespace std;

char a[1000010];
int p[1000010];
void next(int l)
{
    int j=0;
    p[1]=0;
    for(int i=2;i<=l;i++)
    {
        while(j>0&&(a[j+1]!=a[i])) j=p[j];
        if(a[j+1]==a[i]) j+=1;
        p[i]=j;
    }
}
int main()
{
    //freopen("a.txt","r",stdin);
    int n,j=1;
    while(~scanf("%d",&n))
    {
        if(n==0) break;
        printf("Test case #%d
",j++);
        scanf("%s",a+1);
        int l=strlen(a+1);
        next(l);
        for(int i=1;i<=l;i++)
        {
            if(i%(i-p[i])==0&&p[i]!=0)
            {
                printf("%d %d
",i,i/(i-p[i]));
            }
        }
        printf("
");
    }
    return 0;
}