poj 1961 Period

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 13768		Accepted: 6510

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

题目大意：每个测试给定一个字符串，若前i几个字符组成的字符串是周期串，输出i及其周期长度。

 

解题思路：利用kmp算法的next数组，未改进的next数组表示相同前缀后缀的长度，（i-next[i]）可以看以周期，

若有i%(i-next[i])=0，则表示长度为i的该字符串是周期串。

代码如下：

#include <cstdio>
#include <cstring>
const int maxn=1000010;
int a[maxn];
int next[maxn];
int n;
void get_next()
{
    int i=1,j=0;
    next[1]=0;
    while(i<=n)
    {
        if(j==0||a[i]==a[j])
        {
            i++;
            j++;
            next[i]=j;
			if(i%(i-next[i])==0&&a[i]==a[next[i]])
				printf("%d %d\n",i,i/(i-next[i]));	
        }
        else
            j=next[j];
    }
	return ;
}
int main()
{
    int t=0,i;
	char s;
    while(scanf("%d",&n)!=EOF&&n)
    {
        t++;
		memset(next,0,sizeof(next));
		getchar();
		for(i=1;i<=n;i++)
		{
			scanf("%c",&s);
			a[i]=s-'0';
		}
		printf("Test case #%d\n", t);
        get_next();
		printf("\n");
    }
    return 0;
}