(1576)HDU

#include<iostream>
#include<cstdio>
#include<string.h>
#include<cstring>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>

#define LOCAL
#define ll long long
#define lll unsigned long long
#define MAX 1000009
#define eps 1e-8
#define INF 0x7fffffff
#define mod 1000000007

using namespace std;
/*

题意：中文题

想法：扩展GCD。
A%mod = n
推导 n = A - (A/mod)*mod
设A/B = x;
推导 A = Bx
则有 Bx - A/mod*mod = n
设y = A/mod
推导Bx - mody = n
根据 gcd(B,mod) = 1
所以根据extgcd得出  A/B = X = X1*n
*/
int extgcd(int a,int b,int& x,int& y)
{
    if(b==0)
    {
        x= 1,y = 0;
        return a;
    }
    int d = extgcd(b,a%b,x,y);
    int t = x;
    x = y;
    y = t - a/b*y;
    return d;//返回最大公约数
}
int main()
{
    int n,b;
    int x,y;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&b);
        extgcd(b,9973,x,y);
        x = x*n;
        cout<<(x%9973+9973)%9973<<endl;
    }
    return 0;
}