Period kmp

　　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 A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

 

Input

The input file 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

 

这题 需要对next数组具有更深的理解  就很好做了  仔细观察next即可

#include<bits/stdc++.h>
using namespace std;
//input
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);i--)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m);
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define inf 0x3f3f3f3f
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define N 1000000+5
int nex[N];
int lenp;
string p;
void getnext()
{
    lenp=p.size();
    nex[]=-;
    int k=-,j=;
    while(j<lenp)
    {
        if(k==-||p[j]==p[k])
            nex[++j]=++k;
        else k=nex[k];
    }
}

int main()
{
    int n;
    int cas=;
    while(RI(n),n)
    {
        if(!n)break;

        printf("Test case #%d\n",++cas);
        cin>>p;
        getnext();
        rep(i,,lenp)
        {
            if(nex[i]==)continue;
            int j=i-nex[i];
            if(i%j==)
            {
                printf("%d %d\n",i,i/j);
            }
        }
        cout<<endl;
    }
}