POJ - Period(KMP)

题目链接:http://poj.org/problem?id=1961
Time Limit: 3000MS Memory Limit: 30000K

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 AK ,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

Problem solving report:

Description: 给一个字符串，问从串首到那个位置截取的字串是以周期循环的，循环了多少次。
Problem solving: 求完next数组后再从头遍历一遍, i-next[i]就是最小循环节。

Accepted Code:

#include <stdio.h>
using namespace std;
const int MAXN = 1000005;
int nex[MAXN];
char str[MAXN];
void Next(int n) {
    nex[0] = -1;
    int i = 0, j = -1;
    while (i < n) {
        if (~j && str[i] != str[j])
            j = nex[j];
        else nex[++i] = ++j;
    }
}
int main() {
    int n, kase = 0;
    while (~scanf("%d", &n), n) {
        scanf("%s", str);
        Next(n);
        printf("Test case #%d\n", ++kase);
        for (int i = 2; i <= n; i++) {
            int m = i - nex[i];
            if (!(i % m) && i != m)
                printf("%d %d\n", i, i / m);
        }
        printf("\n");
    }
    return 0;
}