UVALive 3026 Period KMP 失配函数处理周期的问题

 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 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
// KMP 失配函数处理周期的问题、 训练指南 P213 
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <ctype.h>
#include <stack>
#include <iostream>
using namespace std;
char P[1000010];
int fail[1000010];
int main()
{
   //  freopen("in.txt","r",stdin);
    int i,j,len;
    int n,num=1;
    while(scanf("%d",&n),n)
    {
        scanf("%s",P);
        fail[0]=fail[1]=0;
        printf("Test case #%d\n",num++);
        for(i=1;i<=n;i++)
         {
             j=fail[i];
             while(j&&P[i]!=P[j]) j=fail[j];
             fail[i+1]=P[i]==P[j]?j+1:0;
             len=i-fail[i];
             if(fail[i]>0&&i%len==0) printf("%d %d\n",i,i/len);
         }
         printf("\n");
    }
    return 0;
}