(A/B)%9973=K

A/B=K+9973*X

A=BK+9973*X*B

A%9973=n;

BK%9973=n;

BK=n+9973*Y

(K/n)*B+(-Y/n)*9973=GCD(B,9973)=1;

求出k/n，求出k

 1 /*
 2     擴展歐幾裏得
 3     擴展歐幾裏德算法是用來在已知a, b求解一組x，y使得ax+by = Gcd(a, b) =d(解一定存在，根據數論中的相關定理)
 4 */
 5 #include<cstdio>
 6 #include<iostream>
 7 #define ll long long
 8 
 9 using
 
 namespace std;
10 
11 ll x,y;
12 void gcd(ll a,ll b)
13 {
14     if(b==0)
15     {
16         x = 1;
17         y = 0;
18         return ;
19     }
20     gcd(b,a%b);
21     ll t = x;
22     x = y;
23     y = t-(a/b)*y;
24 }
25 int main()
26 {
27     int t;
28     int n,b;
29     cin>>t;
30     while
 
(t--)
31     {
32         cin>>n>>b;
33         gcd(b,9973);
34         //x*=n;
35         x = (x%9973+9973)%9973; //最小正數
36         cout<<(x*n)%9973<<endl;
37     }
38     return 0;
39 }

hdu 1576 擴展歐幾裏得