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;
}