Romantic（hdu2699+欧几里德）

最新推荐文章于 2018-07-25 22:49:46 发布

转载最新推荐文章于 2018-07-25 22:49:46 发布 · 139 阅读

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原文链接：http://www.cnblogs.com/yuyixingkong/p/4370240.html

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3218 Accepted Submission(s): 1274

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

Author
yifenfei

题意很简单就是 a * x + b * y = 1 注意的是X必须大于0 SO… a*(x+b) + b*(y-a) = a*x + a*b + b*y - a*b = 1

#include<stdio.h>
#define LL __int64


void exgcd(LL a,LL b,LL& d,LL& x,LL& y)
{
    if(!b)d=a,x=1,y=0;
    else
    {
        exgcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}
int main()
{
    LL a,b;
    while(scanf("%I64d%I64d",&a,&b)!=EOF)
    {
        LL d,x,y;
        exgcd(a,b,d,x,y);
        if(d==1)
        {
            while(x<0)
            {
                //a * x + b * y = 1
                //->  a*(x+b) + b*(y-a) = a*x + a*b + b*y - a*b = 1
                x = x+b;
                y = y-a;
            }
            printf("%I64d %I64d\n",x,y);
        }
        else
            printf("sorry\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/yuyixingkong/p/4370240.html