HDU A/B 扩展欧几里得

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
/*
要求(A/B)%9973
ax+by = c
a*x = c%b 在这里b=9973,
9973*x+B*y = A
B*y = A%9973 在这里求最小的B就是答案
按照求逆元的方法
*/
LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL ans = ex_gcd(b,a%b,x,y);
    LL tmp = x;
    x = y;
    y = tmp - a/b*x;
    return ans;
}
LL cal(LL a,LL b,LL c)
{
    LL x=0,y=0;
    LL gcd = ex_gcd(a,b,x,y);
    if(c%gcd!=0) return -1;
    x *= c/gcd;
    b /= gcd;
    if(b<0) b = -b;
    LL ans = x%b;
    if(ans<0) ans += b;
    return ans;
}
int main()
{
    LL t,n,b;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld%lld",&n,&b);
        printf("%lld\n",cal(b,9973,n));
    }
}