HDOJ 2669 扩展欧几里得

最新推荐文章于 2018-10-06 23:11:40 发布

原创最新推荐文章于 2018-10-06 23:11:40 发布 · 306 阅读

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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，详细讲解了如何找到两个非负整数a和b的解X和Y，使得X*a + Y*b = 1。特别关注于X为非负整数的情况，并提供了具体的实现代码。

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7534 Accepted Submission(s): 3200

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

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lcy | We have carefully selected several similar problems for you: 2668 2674 2671 2670 2672

分析：

裸的扩展欧几里得运用，注意题目要求的是x为正整数。

如果gcd(a,b)!=1 输出sorry

否则如果 x<0, x +=b, y-=a.

x = x0 + b/gcd(a,b)*t , y= y0 -a/gcd(a,b)*t.

代码如下：

#include <cstdio>
using namespace std;

typedef long long ll;
ll a,b,x,y;
ll gcd;

ll extend_gcd(ll a, ll b, ll &x , ll &y){
    if (b==0) {
        x=1; y=0;
        return a;
    }

    ll r = extend_gcd(b,a%b,y,x);
    y -= x*(a/b);
    return r;
}

int main(){
    while (scanf("%lld %lld",&a,&b)!=EOF){
         gcd = extend_gcd(a,b,x,y);
         if (gcd!=1) printf("sorry\n");
         else {
            x %= b;
            if (x<=0) {x+=b; y-=a;}
            printf("%lld %lld\n",x,y);
         }
    }
    return 0;
}