uva 10127

题目链接:uva 10127 - Ones

题目大意:给出n,问说者少要多少为1才干够整除n。

解题思路:等于是高精度取模,直到余数为0为止。

#include <cstdio>
#include <cstring>

int main () {
    int n;
    while (scanf("%d", &n) == 1) {
        int ans = 1, c = 1;
        while (c) {
            c = (c * 10 + 1) % n;
            ans++;
        }
        printf("%d
", ans);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/blfshiye/p/4295181.html