51Nod 1256 乘法逆元(扩展欧几里得)

复制代码

#include <iostream>
#include <algorithm>

using namespace std;
typedef long long LL;

//给予二整数 a 与 b, 必存在有整数 x 与 y 使得ax + by = gcd(a,b)
LL extgcd(LL a, LL b, LL &x, LL &y){
    LL d = a;
    if (b != 0){
        d = extgcd(b, a%b, y, x);
        y -= (a / b)*x;
    }
    else{
        x = 1;
        y = 0;
    }
    return d;
}

LL inv(LL a, LL m)
{
    LL x, y;
    extgcd(a, m, x, y);
    return (m + x%m) % m;
}

int main(){
    ios::sync_with_stdio(false);
    LL n, m;
    while (cin >> m >> n)
    {
        cout << inv(m, n) << endl;
    }
    return 0;
}