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.

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.

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.

3
aaa
12
aabaabaabaab
0

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

JGShining

 

本题要求所给字符串的长度大于等于2的可以由某一子串循环得到的前缀的长度和循环次数。

此题首先得求最小循环节，用KMP算法。

 

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int MAXN=1000000+100;
char str[MAXN];//W为模式串，T为主串
int next[MAXN];

//KMP算法中计算next[]数组
void getNext(char *p)
{
    int j,k,len=strlen(p);
    j=0;
    k=-1;
    next[0]=-1;
    while(j<len)
    {
        if(k==-1||p[j]==p[k])
        {
            next[++j]=++k;
        }
        else k=next[k];
    }
}

int main()
{
	int n,i,tag=1;
	while(~scanf("%d",&n),n)
	{
		scanf("%s",str);
		getNext(str);
		printf("Test case #%d\n",tag++);
		for(i=2;i<=n;i++)
		{
			if(next[i]&&i%(i-next[i])==0)
				printf("%d %d\n",i,i/(i-next[i]));
		}
		printf("\n");
	}
	return 0;
}