J - Modular Inverse ZOJ - 3609

The modular modular multiplicative inverse of an integer a modulo m is an integer x such that . This is equivalent to . a-1≡x (mod m)ax≡1 (mod m)

Input

There are multiple test cases. The first line of input is an integer T ≈ 2000 indicating the number of test cases.

Each test case contains two integers 0 < a ≤ 1000 and 0 < m ≤ 1000.

Output

For each test case, output the smallest positive x. If such x doesn't exist, output "Not Exist".

Sample Input

3
3 11
4 12
5 13

Sample Output

4
Not Exist
8

References

• http://en.wikipedia.org/wiki/Modular_inverse

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
/*
It's really too yin, it's still my own inertia, too strong, including myself not very familiar with Ni Yuan
So it is directly very confusing
Even when it was 1, you can mod at the same time on both sides and serve it.
*/

intmain()
{
	ios::sync_with_stdio(false);
	int T;
	cin>>T;
	while(T--){
		ll a,m,ans;
		cin>>a>>m;
		int flag=0;
		for(ll i=1;i<=m;i++){
			if((a*i)%m==1%m){
				flag=1;
				ans=i;
				break;
			}
		}
		if(flag)
			cout<<ans<<endl;
		else
			cout<<"Not Exist"<<endl;
	}
	return 0;
}

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

ll exgcd(ll a,ll b,ll& d,ll& x,ll&y){
	if(!b){  
        d=a;x=1;y=0;  
    }  
    else{  
    	exgcd(b,a%b,d,y,x);  
        y-=x*(a/b);  
    }
}
/*
If you turn this off, don't use scanf, otherwise it will be wa
*/

intmain()
{
	ios::sync_with_stdio(false);
	int T;
	cin>>T;
	while(T--){
		ll a,m;
		cin>>a>>m;
		if(m==1){
			cout<<"1"<<endl;
			continue;
		}
		ll x,y,gcd;
		exgcd(a,m,gcd,x,y);
		if(gcd!=1){
			cout<<"Not Exist"<<endl;
			continue;
		}
		if(x<=0){
			x=(x%m+m)%m;
		}
		cout<<x<<endl;
	}
	return 0;
}