hdu 2669 Romantic（线性同余，扩展欧几里得）

扩展欧几里得算法详解

最新推荐文章于 2019-02-10 20:59:51 发布

原创最新推荐文章于 2019-02-10 20:59:51 发布 · 942 阅读

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#ACM #HDU #数论 #扩展欧几里得

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本文介绍了一道关于扩展欧几里得算法的题目，要求对于给定的两个正整数a和b，找到非负整数X和整数Y使得X*a + Y*b = 1成立。若无解，则输出特定提示。文章提供了详细的解题思路及代码实现。

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4433 Accepted Submission(s): 1868

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

题意:给你a，b，求ax+by=1的解，如果有多解输出x（非负数）最小的解。

思路：扩展欧几里得模板题~不多说了

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define LL long long
LL extend_gcd(LL a,LL b,LL &x,LL &y)
{
    if(!b)
    {
        x=1,y=0;
        return a;
    }
    LL d=extend_gcd(b,a%b,x,y);
    LL x1=x;
    x=y;
    y=x1-(a/b)*y;
    return d;
}
int main()
{
    LL a,b;
    while(~scanf("%lld %lld",&a,&b))
    {
        LL x,y;
        LL d=extend_gcd(a,b,x,y);
        if(1%d!=0) printf("sorry\n");
        else
        {
            x=x/d;
            LL mod=b/d;
            x=(x%mod+mod)%mod;
            LL y=(1-a*x)/b;
            printf("%lld %lld\n",x,y);
        }
    }
    return 0;
}