POJ_1961

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 19817		Accepted: 9640

Description

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 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

引用大佬的博客:简单的对next数组的应用，http://www.cnblogs.com/jackge/archive/2013/01/05/2846006.html
以及KMP算法的详细解释:https://blog.csdn.net/v_july_v/article/details/7041827

看了大佬的博客们写的代码:

#include<stdio.h>

int next[1000005];
int get_next(int n,char str[]){
	int i=0,j=-1;
	next[0]=-1;
	//当循环到没有正确匹配的str时都令j=next[j]
	//即上一次匹配到的最近 正确的 j 的下一个
	//KMP算法，这样可以减少消耗
	while(i<n){
		if(j==-1||str[i]==str[j]){
			i++;
			j++;
			next[i]=j;
		}else{
			j=next[j];
		}
	}
}
int main(){
	int i,length,k=1;
	int temp;
	char str[1000005];
	int count=1;
	while(scanf("%d",&length)&&length){
		scanf("%s",str);
		get_next(length,str);
		printf("Test case #%d\n",count);
		count++;
		for(i=1;i<=length;i++){
			temp=i-next[i];
			//temp为最短循环节的长度
			if(i%temp==0&&i/temp>1)
				printf("%d %d\n",i,i/temp);
			//i%temp==0 是因为最小循环节需要被字符串长度整除
			//同时 需要被整除
		}
		printf("\n");
	}
	return 0;
}