HDU 2669 Romantic【扩展欧几里得】-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_37867156/article/details/82934128

本文详细介绍了扩展欧几里得算法的应用，通过解决特定的数学问题，展示了如何找到两个非负整数的倍数组合等于1的解。文章包含了一个C++实现的示例代码，用于解释算法的步骤和输出格式。

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The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei

Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.

Input

The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)

Output

output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.

Sample Input

77 51
10 44
34 79

Sample Output

2 -3
sorry
7 -3

题解：

扩展欧几里得定理：对于任意整数 a, b，都存在 x, y 使得 ax + by = gcd(a, b)。

对于一般的方程 ax + by = c，它有解当且仅当 d|c，而本题的 c 等于1，所以d也要等于1才有解。无解输出 “sorry”。并且要求的解 x 必须大于等于0，所以又需要用到通解公式： x = (c/d)x + (b/d)k， y = (c/d)y - (a/d)k, k 为任意整数。

#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;

int exgcd(int a, int b, int &x, int &y){
    if(b == 0){
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a%b, x, y);
    int z = x;
    x = y, y = z - (a/b) * y;
    return d;
}
int main()
{
    int a, b;
    while(cin >> a >> b){
        int x, y;
        int d = exgcd(a, b, x ,y);
        if(d != 1){
            cout << "sorry" << endl;
        }else{
            while(x < 0){
                x += b;
                y -= a;
            }
            cout << x << " " << y << endl;
        }
    }
    return 0;
}