Euclid Problem - PC110703

weixin_30639719發表於2020-04-05

歡迎訪問我的新部落格：http://www.milkcu.com/blog/

原文地址：http://www.milkcu.com/blog/archives/uva10104.html

原創：Euclid Problem - PC110703

作者：MilkCu

題目描述


 Euclid Problem 

The Problem

From Euclid it is known that for any positive integers A and Bthere exist such integersX andY that AX+BY=D,where D is the greatest common divisor ofA andB.The problem is to find for given A and B correspondingX,Y and D.

The Input

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

The Output

For each input line the output line should consist of threeintegers X, Y andD, separated with space. If there are severalsuchX and Y, you should output that pair for which|X|+|Y| is the minimal (primarily) andX<=Y (secondarily).

Sample Input

4 6
17 17

Sample Output

-1 1 2
0 1 17

解題思路

本題考查的是求解最大公約數（greatest common divisor，也稱gcd）的歐幾里得（Euclid）演算法。

Euclid演算法建立在兩個事實的基礎上：
1. 如果b|a（b整除a），則gcd(a, b) = b；
2. 如果存在整數t和r，使得a = bt + r，則gcd(a, b) = gcd(b, r)。

Euclid演算法是遞迴的，它不停的把較大的整數替換為它除以較小整數的餘數。

Euclid演算法指出，存在x和y，使
a * x + b * y = gcd(a, b)
又有
gcd(a, b) = gcd(b, a % b)
為了方便計算，我們將上式寫為
gcd(a, b) = gcd(b, a - b * floor(a / b))
假設我們已經找到整數x'和y'，使得
b * x' + (a - b * floor(a / b)) * y' = gcd(a, b)
整理上式，得到
a * y' + b * (x' - b * floor(a / b) * y') = gcd(a, b)
與a * x + b * y = gcd(a, b)相對應，得到
x = y'
y = x' - floor(a / b) * y'

程式碼實現

#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int gcd(int a, int b, int & x, int & y) {
	int x1, y1;
	if(a < b) {
		return gcd(b, a, y, x);
	}
	if(b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	int g = gcd(b, a % b, x1, y1);
	x = y1;
	y = x1 - floor(a / b) * y1;
	return g;
}
int main(void) {
	int a, b;
	while(cin >> a >> b) {
		int x, y;
		int g = gcd(a, b, x, y);
		cout << x << " " << y << " " << g << endl;
	}
	return 0;
}

（全文完）

本文地址：http://blog.csdn.net/milkcu/article/details/23590217

轉載於:https://www.cnblogs.com/milkcu/p/3808850.html

相關文章