hdu 2669 (Romantic)扩展偶几里德

学到现在发现自己是会的越来越少了，不会的越来越多了！

本体给出一个方程a*x+b*y=1;

给出a，b的值，求出x，y，其中y为负值。就是求解直线方程。

思路：求出a，b的最大公约数gcd（a，b）；方程左边（一定能够整除最大公约数）除于最大公约数肯定为一个常数，所以方程右边只有对这个最大共约能整除的时候两边才有可能相等。

具体的实现过程还有待研究。菜菜，还有很多疑问啊！

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

long long exgcd(long long a,long long b,long long &x,long long &y)

{



    if(b==0)  {   //？

        x=1;y=0;

        return a;

    }

    else{

    long long d=exgcd(b,a%b,x,y);

    long long temp=x; //？

    x=y;

    y=temp-a/b*y;

    return d;//返回最大共公约数

    }



}

int main()

{



    long long x,y,a,b;

    while(cin>>a>>b)

    {

        long long ans=exgcd(a,b,x,y);

        if(1%ans)//如果方程右边的数不能够整除最大公约数

            printf("sorry\n");

        else{

            while(x<0)//？

                x+=b,y-=a;

            cout<<x<<" "<<y<<endl;

        }





    }

    return 0;

}

 

 

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

77 51

10 44

34 79

 

 

Sample Output

2 -3

sorry

7 -3