hdoj 2669 Romantic 【扩展欧几里得求解最小非负解】

最新推荐文章于 2021-01-27 16:26:58 发布

Big_Rui

最新推荐文章于 2021-01-27 16:26:58 发布

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分类专栏：扩展欧几里得

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本文介绍了一种利用扩展欧几里得算法求解特定线性方程的方法，特别是当给定两个正整数a和b时，如何找到非负整数X和整数Y使得X*a + Y*b = 1成立。如果不存在这样的解，则输出'sorry'。

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题意：给定a,b,求想x，y使a*x+b*y=1;

思路；扩展欧几里得求乘法逆元，x好求，主要是y，直接y=1-a*x;

一开始想到了x=x0+b*t;y=y0-a*t;算t，发现比较麻烦，还是直接算来的方便

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;

ll egcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0){
        x=1;y=0;
        return a;
    }
    ll d=egcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}

int main()
{
    ll a,b,x,y,d,t;
    while(~scanf("%lld%lld",&a,&b))
    {

        d=egcd(a,b,x,y);
        if(d!=1) {printf("sorry\n");continue;}
        x=((x%b)+b)%b;
        y=(1-a*x)/b;
        printf("%lld %lld\n",x,y);
    }
}

题目：

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