HDU 2669 （扩展欧几里德算法）

最新推荐文章于 2020-02-03 00:07:00 发布

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最新推荐文章于 2020-02-03 00:07:00 发布

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本文深入探讨了扩展欧几里得算法的应用，通过解决特定数学问题，寻找两个非负整数a和b的非负整数解X和Y，使等式Xa+Yb=1成立。详细讲解了算法原理，包括如何判断解的存在性和求解过程，并提供了代码实现。

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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 Xa + Yb = 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

由扩展欧几里得定理有，如果ax+by = d 要使得方程有解必有gcd(a,b)为d的约数。
而此题的d = 1 所以若gcd(a,b)!=1，则应该输出sorry
注意，输出的x应为最小的非负整数，这就需要用到x，y所有解的公式：
x，y所有解：
假设d=gcd(a,b). 那么x=x0+b/dt; y=y0-a/dt;其中t为任意常整数

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define inf 1000000
using namespace std;
ll exgcd(ll a,ll b,ll &x,ll &y){
    int d=a;
    if(b!=0){
        d=exgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }else{
        x=1;y=0;
    }
    return d;
}
int main(){
    ll a,b,x,y,g;
    while(scanf("%lld %lld",&a,&b)!=EOF){
        g=exgcd(a,b,x,y);
        if(g!=1){
            printf("sorry\n");
        }else{
            if(x<0){
                x=x+b;
                y=y-a;
            }
            printf("%lld %lld\n",x,y);
        }
    }
    return 0;
}