c - 为什么我在这个问题上得到了错误的答案(Uva OJ 455)

Question

我很喜欢this problem (乌瓦455):

A character string is said to have period k if it can be formed by concatenating one or more repetitions of another string of length k. For example, the string ”abcabcabcabc” has period 3, since it is formed by 4 repetitions of the string ”abc”. It also has periods 6 (two repetitions of ”abcabc”) and 12 (one repetition of ”abcabcabcabc”).

Write a program to read a character string and determine its smallest period.

Input

The first line oif the input file will contain a single integer N indicating how many test case that your program will test followed by a blank line. Each test case will contain a single character string of up to 80 non-blank characters. Two consecutive input will separated by a blank line.

Output

An integer denoting the smallest period of the input string for each input. Two consecutive output are separated by a blank line.

Sample Input

1

HoHoHo

Sample Output

2

我已经检查了我能想象到的所有测试用例，并且所有测试用例都返回了正确的结果，但我仍然在在线判断中得到错误的答案。我哪里做错了？(英语不是我的母语；请原谅打字或语法错误。)

#include <stdio.h>
#include <string.h>
#define maxn 85

int check(char* s, int per){
    for(int i = 0; i < strlen(s) - per; i++){
        if(s[i + per] != s[i])  return 0;
    }
    return 1;
}

int main(){
    int T;
    scanf("%d", &T);
    char s[maxn];
    while(T--){
        scanf("%s", s);
        int len = strlen(s);
        bool OK = false;
    for(int i = 1; i <= len/2 && (len % i == 0); i++){//That's wrong.
        if(check(s, i)){
            printf("%d\n", i);
        OK = true;
        break;
        }
    }
        if(!OK) printf("%d\n", len);
        if(T)   printf("\n");
    }
    return 0;
}

Answer 1

问题出在 for(int i = 1; i <= len/2 && (len % i == 0); i++) 。一遇到i就停下来不能除len ，而不是跳过它。

将循环写为:

for (int i = 1; i <= len/2; i++) {
    if (len % i != 0) continue;

    ...
}