【ZOJ 3609】Modular Inverse 最小乘法逆元

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

题解：
最小乘法逆元：由ax≡1 (mod m)得：转化为解线性方程ax+by=1
需要注意的地方：最小解取模时不能写成（x%t+t）%t 因为此题要的是正数解 这样写有时会输出0

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(!b)
    {
        x=1;y=0;
        return a;
    }
    ll r=exgcd(b,a%b,x,y);
    ll t=x;
    x=y;
    y=t-a/b*y;
    return r;
}
void work(ll a,ll b,ll c)
{
    ll x,y;
    ll r=exgcd(a,b,x,y);
    if(c%r!=0){
        printf("Not Exist
");
        return ;
    }
    x*=c/r;
    ll t=b/r;
    if(t<0)t=-t;
    x%=t;
    if(x<=0)x+=t;
    printf("%lld
",x);
}
int main()
{
    int T;
    ll a,b;
    scanf("%d",&T);
    while(T--){
        scanf("%lld%lld",&a,&b);
        work(a,b,1);
    }
    return 0;
}