HDU2669：Romantic（扩展欧几里得）

最新推荐文章于 2022-11-17 19:16:18 发布

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本文介绍了一个基于扩展欧几里得算法的数学问题求解案例，该问题要求找到两个非负整数a和b的解X和Y，使得X*a + Y*b = 1。若不存在解，则输出'sorry'。文章提供了完整的C++实现代码，并通过样例展示了算法的应用。

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6016 Accepted Submission(s): 2563

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

思路：裸扩展欧几里得，注意x不为负数，long long用I64输入输出。

# include <iostream>
# include <cstdio>
# define LL long long
using namespace std;
LL x, y, d;

LL ex_gcd(LL a, LL b)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    d = ex_gcd(b, a%b);
    LL t = x;
    x = y;
    y = t-(a/b)*y;
    return d;
}

int main()
{
    LL a, b;
    while(~scanf("%lld%lld",&a,&b))
    {
        d = ex_gcd(a, b);
        if(d != 1)
        {
            puts("sorry");
            continue;
        }
        LL k = x/b;
        if(x-b*k < 0)
            --k;
        printf("%I64d %I64d\n",x-b*k, y+a*k);

    }
    return 0;
}