HDU2669-Romantic

最新推荐文章于 2019-09-03 10:36:03 发布

原创最新推荐文章于 2019-09-03 10:36:03 发布 · 380 阅读

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本文介绍了一个基于扩展欧几里德算法的数学问题解决方法。通过给出的两个正整数a和b，找到满足特定等式的最小非负整数X及相应的整数Y。文章提供了详细的解题思路与C++实现代码。

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Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5236 Accepted Submission(s): 2204

Problem Description

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

Sample Output

  
   2 -3
sorry
7 -3

Author

yifenfei

Source

HDU女生专场公开赛——谁说女子不如男

题意：找出最小的非负整数X，使之满足式子x*a+y*b=1。

解题思路：简单的扩展欧几里德。

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

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

int main()
{
    int a,b,x,y;
    while(~scanf("%d %d",&a,&b))
    {
        int d=exgcd(a,b,x,y);
        if(d!=1)
            printf("sorry\n");
        else
        {
            while(x<0)
            {
                x+=b;
                y-=a;
            }
            printf("%d %d\n",x,y);
        }
    }
    return 0;
}