LightOJ 1078 Integer Divisibility (同余定理)

Integer Divisibility
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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Description
If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.

For example you have to find a multiple of 3 which contains only 1’s. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3’s then, the result is 6, because 333333 is divisible by 7.

Input
Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case will contain two integers n (0 < n ≤ 106 and n will not be divisible by 2 or 5) and the allowable digit (1 ≤ digit ≤ 9).

Output
For each case, print the case number and the number of digits of such multiple. If several solutions are there; report the minimum one.

Sample Input
3
3 1
7 3
9901 1
Sample Output
Case 1: 3
Case 2: 6
Case 3: 12

同余定理， 比如第一个，（100 + 10 + 1） % 3 = （100 % 3 + 10 % 3 + 1 % 3）% 3， 题解如下

#include <cstdio>#include <cstring>int main(){    int t, n, m, ans, wn, num;    scanf("%d", &t);    for(int ca=1; ca<=t; ca++)    {        scanf("%d%d", &n, &m);        num = 1;        ans = m % n;        while(ans)        {            num++;            ans = (ans * 10 % n + m % n) % n;        }        printf("Case %d: %d\n", ca, num);    }    return 0;}