Modular Inverse（逆元）

题目连接：https://cn.vjudge.net/problem/ZOJ-3609

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll ex_gcd(ll a,ll b,ll &x,ll &y)
{
    if(a==0&&b==0)return -1;
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    ll d=ex_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
ll ex_gcd1(ll a,ll n)
{
    ll x,y;
    ll d=ex_gcd(a,n,x,y);
    if(d==1)
    {
        if(x<0)
            return (x%n+n)%n;
        else if(x==0)
            return n;
        return x;
    }
    else return -1;
}
ll gcd(ll a,ll b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll a,b;
        scanf("%lld%lld",&a,&b);
        if(gcd(a,b)!=1)
        {
            printf("Not Exist\n");
            continue;
        }
        ll g=ex_gcd1(a,b);
        if(g==-1)
            printf("Not Exist\n");
        else
        {
            printf("%lld\n",g);
        }
    }
    return 0;
}