51Nod 1256 乘法逆元(扩展欧几里得)

 1 #include <iostream>
 2 #include <algorithm>
 3
 4 using namespace std;
 5 typedef long long LL;
 6
 7 //给予二整数 a 与 b, 必存在有整数 x 与 y 使得ax + by = gcd(a,b)
 8 LL extgcd(LL a, LL b, LL &x, LL &y){
 9     LL d = a;
10     if (b != 0){
11         d = extgcd(b, a%b, y, x);
12         y -= (a / b)*x;
13     }
14     else{
15         x = 1;
16         y = 0;
17     }
18     return d;
19 }
20
21 LL inv(LL a, LL m)
22 {
23     LL x, y;
24     extgcd(a, m, x, y);
25     return (m + x%m) % m;
26 }
27
28 int main(){
29     ios::sync_with_stdio(false);
30     LL n, m;
31     while (cin >> m >> n)
32     {
33         cout << inv(m, n) << endl;
34     }
35     return 0;
36 }

原文地址：https://www.cnblogs.com/jaydenouyang/p/8721235.html