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

缺乏、安全感 2022-05-18 00:05 189阅读 0赞

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;
    }