hdu 2669(扩展欧几里得)

最新推荐文章于 2025-09-10 22:21:01 发布

转载最新推荐文章于 2025-09-10 22:21:01 发布 · 67 阅读

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原文链接：http://www.cnblogs.com/liyinggang/p/5528236.html

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#java

本文介绍了一种通过扩展欧几里得算法解决特定数学问题的方法：对于给定的两个非负整数a和b，找到非负整数X和整数Y，使得X*a + Y*b = 1成立。如果不存在这样的解，则输出sorry。文章提供了完整的C++代码实现。

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4400 Accepted Submission(s): 1852

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

51
44
79

Sample Output

2 -3
sorry
7 -3

题解:由ax+by = gcd(a,b) 当 a,b互素是才会有解。然后X要是尽量小的正数，假设我们得到的是 X0 ，在有解的情况下方程ax+by=c的通解为 {X0*(c/d)+k*b/gcd(a,b)} (k = ..-2,-1,0,1,2..)

所以我们可以得到在最小的X为 (X0%b+b)%b.代入得Y

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long LL;

LL extend_gcd(LL a,LL b,LL &x,LL &y){
    if(!b){
        x=1,y = 0;
        return a;
    }else{
        LL x1,y1;
        LL d = extend_gcd(b,a%b,x1,y1);
        x = y1;
        y = x1 - a/b*y1;
        return d;
    }
}
int main()
{
    LL a,b,x,y;
    while(~scanf("%lld%lld",&a,&b)){
        LL d = extend_gcd(a,b,x,y);
        if(d!=1){
            printf("sorry\n");
        }else{
            x = (x%b+b)%b;
            y = (1-a*x)/b;
            printf("%lld %lld\n",x,y);
        }
    }
    return 0;
}
 