同余定理
Light OJ---1078
题意: 给你两个整数,n和m,问有最少有几个m组成的数能被n整除。。。。。不用同余会超时。。。。
思路:同余定理:
(A + B) mod M = ( A mod M + B mod M ) mod M
(A * B) mod M = ((A mod M) *( B mod M)) mod M
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdio> #include <cstring> #include<iostream> #include<queue> #include<stack> #include<cmath> #include <map> #include <algorithm> using namespace std; map<string, string>mp; vector<vector<int> >G; const int INF=0x3f3f3f3f; const int maxn=1100; int main() { int T, cas=1; scanf("%d", &T); while(T--) { int n, m; scanf("%d %d", &n, &m); int t=m%n; int ans=1; while(t) { t=(t*10+m)%n; ans++; } printf("Case %d: %d\n", cas++, ans); } return 0; }