数论：拓展欧几里得算法

算法描述就是:

求整数 x和y 使得 ax + by = 1.

可以发现, 如果gcd(a, b) ≠ 1，则显然无解.

反之, 如果gcd(a, b) = 1, 则可以通过拓展原来的 辗转相除法 来求解.

事实上,一定存在整数对(x, y)使得ax+by = gcd(a,b) = 1

代码如下:

#include <iostream>
using namespace std;

int extgcd(int a, int b, int& x, int& y)
{
    int d = a;
    if (b != 0)
    {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}

void solve()
{
    int x = 0, y = 0;
    int a, b;
    cin >> a >> b;
    int d = extgcd(a, b, x, y);
    cout << d << "=" << a << "*" << x << "+" << b << "*" << y << endl; 
}

int main()
{
    solve();
    
    return 0;
}

 

题解：直接套用模板 

#include <iostream>
#include <cstdio>
using namespace std;

int extgcd(int a, int b, int& x, int& y)
{
    int d = a;
    if (b != 0)
    {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}

void solve()
{
    int a = 97, b = 127;
    int x = 0, y = 0;
    int d = extgcd(a, b, x, y);
    cout << d << "=" << a << "*" << x << "+" << b << "*" << y << endl;
}

int main()
{
    solve();
    
    return 0;
}