poj1961 KMP（循环节）

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Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 17696 Accepted: 8523

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

3aaa12aabaabaabaab0

Sample Output

Test case #12 23 3Test case #22 26 29 312 4
给你一个字符串，求这个字符串到第i个字符为止的循环节的次数。
比如aabaabaabaab，长度为12.到第二个a时，a出现2次，输出2.到第二个b时，aab出现了2次，输出2.到第三个b时，aab出现3次，输出3.到第四个b时，aab出现4次，输出4.
遍历一遍即可，注意理解题意
#include<cstdio>#include<cstring>using namespace std;#define EPS 1e-7#define clr(x) memset(x, 0, sizeof(x))#define long long ll#define double db#define PI acos(-1.0)const int INF = 0x3f3f3f3f;const int MOD=1000000007;const int MAXN = 1000000 + 5;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, 1, -1};int next[MAXN];char str[MAXN];void kmp_pre(char x[], int m){    int i = 0, j = next[0] = -1;    while(i != m)    {        if(j == -1 || str[i] == str[j])            next[++i] = ++j;        else            j = next[j];    }}int main(){    int n;    int cnt = 0;    while(~scanf("%d", &n) && n)    {        scanf("%s", str);        kmp_pre(str, n);        printf("Test case #%d\n", ++cnt);        int ans;        for(int i = 1; i <= n; ++i)        {            ans = i - next[i];            if(i != ans && i % ans == 0)                printf("%d %d\n", i, i / ans);        }        puts("");    }    return 0;}

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