Integer Divisibility LightOJ - 1078——同余定理+逆元（大数取模）

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

---

emmm，暴力就行，一定要记得取模

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long ll;

ll pow_mod(ll a,ll b,ll c)
{
    ll ans=1;
    a%=c;
    while(b)
    {
        if(b&1) ans=ans*a%c;
        a=a*a%c;
        b>>=1;
    }
    return ans;
 } 

ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    ll r=exgcd(b,a%b,y,x);
    y-=a/b*x;
    return r;
}

ll inv(ll a,ll mod)
{
    ll x,y;
    ll d=exgcd(a,mod,x,y);
    if(d==1)    return (x%mod+mod)%mod;
    return -1;
}

int main()
{
    int t,n,m;
    scanf("%d",&t);
    int c=t;
    while(t--)
    {
        scanf("%d %d",&n,&m);
        int sum=m%n;
        int ans=1;
        while(sum%=n)
        {
            sum=(sum*10+m)%n;
            ans++;  
        }       
        printf("Case %d: %d\n",c-t,ans);
    }
    return 0;
}