hdu 1358 Period【KMP字符串循环】

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
题意：
给一个字符串，问从首位到i位截取的字符串中的一个子串最多循环了多少次，输出i位和循环次数（大于1）；
思路：异曲同工的题目
类似与之前的博客，只不过遍历一遍，主要是 ：i %（i - f[i]）== 0 且 大于1。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define max_n 1000010using namespace std;char str1[max_n], str2[max_n];int f[max_n];int n;void getnext() {    int i = 0, j = -1;    f[0] = -1;    while(i < n) {        if(j == -1 || str2[j] == str2[i])            f[++i] = ++j;        else             j = f[j];    }}void solve() {    for(int i = 1; i <= n; i++) {        int res = i - f[i];        if(i % res == 0 && i / res > 1) {            printf("%d %d\n", i, i / res);        }    }}int main() {    while(scanf("%d", &n) && n) {        scanf("%s", str2);        static int p  = 1;        if(p > 1) printf("\n");        printf("Test case #%d\n", p++);        getnext();        solve();    }    return 0;}