POJ1961 Period (kmp) 题解

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 16462		Accepted: 7903

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 OutputTest case #1

2 2
3 3

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


题目链接：http://poj.org/problem?id=1961

把主串的每一种前缀当作小主串，判断小主串是否为子串的叠加，如果len-next[len]能被len整除，则该小主串是由长度为len-next[len]的子串len/(len-next[len])次叠加组成的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxs = 1000010;
int Next[maxs];
char s[maxs];
int N;
void get_next()
{
    int i = 0, j = -1;
    Next[0] = -1;
    while(i < N)
    {
        if(j == -1 || s[i] == s[j])
        {
            i++;
            j++;
            Next[i] = j;
        }
        else
            j = Next[j];
    }
}
int main()
{
    //freopen("poj1961.txt","r",stdin);
    int k = 0;
    while(~scanf("%d",&N) && N != 0)
    {
        scanf("%s",s);
        k++;
        get_next();
        cout << "Test case #" << k << endl;
        for(int i = 1; i <= N; i++)
        {
            if(Next[i] != 0 && i%(i-Next[i]) == 0)
                cout << i << ' ' << i/(i-Next[i]) << endl;
        }
        cout << endl;
    }
}