UVA 1328

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=36131

题意：给出一个长度为n的字符串，要求找到一些i，满足说从1~i为多个个的重复子串构成，并输出子串的个数。

题解：对kmp中预处理的数组的理解   

//作者:1085422276
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include<bits/stdc++.h>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
const int inf = 10000000;
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    ll temp,p;
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    p=exgcd(b,a%b,x,y);
    temp=x;
    x=y;
    y=temp-(a/b)*y;
    return p;
}
//*******************************
int n,p[1000001];char a[1000001];
int main()
{
    int oo=1;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)break;
        scanf("%s",a+1);
        memset(p,0,sizeof(p));
        int j=0;
        for(int i=2;i<=n;i++)
        {
            while(j>0&&a[j+1]!=a[i])j=p[j];
            if(a[j+1]==a[i])j++;
            p[i]=j;
        }
        cout<<"Test case #"<<oo++<<endl;
        for(int i=2;i<=n;i++)
        {
            if(p[i]>0&&i%(i-p[i])==0){
                cout<<i<<" "<<i/(i-p[i])<<endl;
            }
        }
        cout<<endl;
    }
    return 0;
}