UVA10104 Euclid Problem【数论 扩展欧几里得算法】

题目链接:10104 - Euclid Problem

From Euclid it is known that for any positive integers A and B there exist such integers X and Y that AX + BY = D, where D is the greatest common divisor of A and B. The problem is to find for given A and B corresponding X, Y and D.

Input

The input will consist of a set of lines with the integer numbers A and B, separated with space (A, B < 1000000001).

Output

For each input line the output line should consist of three integers X, Y and D, separated with space. If there are several such X and Y , you should output that pair for which |X| + |Y | is the minimal. If there are several X and Y satisfying the minimal criteria, output the pair for which X ≤ Y .

Sample Input

4 6

17 17

Sample Output

-1 1 2

0 1 17

 

题意简述:

求出最大公约数gcd,并求解出使得: a*x + b*y = gcd 的通解 x 和 y。

(求通解x,y便是扩展欧几里得算法和欧几里得算法的区别所在)

 

题记:

此题为扩展欧几里得算法的裸题,记住这个套路,以后遇到直接用。

 

C++语言程序如下:

/* UVA10104 Euclid Problem */
 
#include <iostream>
 
using namespace std;
 
// 扩展欧几里得算法
int exgcd(int a, int b, int &x, int &y)
{
    if (a % b == 0) {
        x = 0;
        y = 1;
        return b;
    }else {
        int r, tx, ty;
        r = exgcd(b, a % b, tx, ty);
        x = ty;
        y = tx - a / b * ty;
        return r;
    }
}
 
int main()
{
    int a, b, r, x, y;
 
    while(cin >> a >> b) {
        r = exgcd(a, b, x, y);
 
        cout << x << " " << y << " " << r << endl;
    }
 
    return 0;
}

 

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转载自blog.csdn.net/fyy_lufan/article/details/81158742
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