HDU 2669 Romantic（拓展欧几里得）

最新推荐文章于 2024-09-25 22:03:46 发布

原创最新推荐文章于 2024-09-25 22:03:46 发布 · 354 阅读

0 ·

CC 4.0 BY-SA版权

HDU 专栏收录该内容

170 篇文章

订阅专栏

本文介绍了一道数学问题的解决方法，通过拓展欧几里得算法寻找两个非负整数a和b的整数解X和Y，使得X*a + Y*b = 1。文章提供了详细的算法实现步骤及C++代码示例。

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6585 Accepted Submission(s): 2757

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：

X*a + Y*b = 1.给你a b，求x y，x要最小非负数。

point：

拓展欧几里得：

求a * x + b * y = c的整数解

先计算Gcd(a,b)，若c不能被Gcd(a,b)整除，则方程无整数解；否则，在方程两边同时除以Gcd(a,b)，得到新的不定方程a' * x + b' * y = c'

x = c' * x0 + b' * t

y = c' * y0 - a' * t

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define ll long long
int exgcd(int a,int b,int &x,int &y)//ax+by=gcd(a,b);
{
    int d=a;
    if(b!=0)
    {
        d=exgcd(b,a%b,y,x);
        y=y-(a/b)*x;
    }
    else
    {
        x=1;
        y=0;
    }
    return d;
}
int main()
{
    int a,b;
    while(~scanf("%d %d",&a,&b))
    {
        int flag=0;
        if(a<b) swap(a, b),flag=1;
        int x,y;
        int d=exgcd(a,b,x,y);
        if(d!=1) printf("sorry\n");
        else
        {
            if(flag)
            {
                if(y<0)
                {
                    y=y+a/d;
                    x=x-b/d;
                }
                printf("%d %d\n",y,x);
            }
            else
            {
                if(x<0)
                {
                    x=x+b/d;
                    y=y-a/d;
                }
                printf("%d %d\n",x,y);
            }
        }
        
    }
    return 0;
}