hdu-2669(扩展欧几里得）

太过爱你忘了你带给我的痛 2022-06-10 08:48 103阅读 0赞

Problem Description

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

题目题意十分明显，就是让你求x\*a+y\*b=1,这个不定方程，x的解有一个条件就是，最小的正整数。

题目分析:扩展欧几里得算法可以解决这样的不定方程x\*a+y\*b=c，对于这样的一个方程.

我们依据的是最初的方程a\*x+b\*y=gcd(a,b) ，当c|gcd(a,b)时方程有解。

bool linear_equation(int a,int b,int c,int &x,int &y)
    {
        int d=exgcd(a,b,x,y);
        if(c%d)//未满足要求
            return false;
        int k=c/d;
        x*=k; y*=k;    //求得的只是其中一组解
        return true;
    }
    ax+by=c其他的解满足:
    X=x1+b/gcd(a,b)*t;//方程所有的解
    Y=y1-a/gcd(a,b)*t;
    其中t为任意整数

代码如下:

#include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    
    int gcd(int a,int b)
    {
        if (b==0) return a;
        else gcd(b,a%b);
    }
    int exgcd(int a,int b,int &x,int &y)//解决ax+by=gcd(a,b)
    {
        if (b==0) {
            x=1;
            y=0;
            return a;
        }
        int r=exgcd(b,a%b,x,y);
        int t=x;
        x=y;
        y=t-a/b*y;
        return r;
    }
    
    bool linear_equation(int a,int b,int c,int &x,int &y)
    {
        int d=exgcd(a,b,x,y);
        if (c%d)
            return false;
        int k=c/d;
        x*=k,y*=k;
        int t=0;
        int temp=b/d;
        if (x<0) {//求最小的整数x
            while (x<0) {
                t++;
                x=x+temp*t;
            }
            printf("%d %d\n",x,y-a/d*t);
        }
        else {
            while (x>0) {
                t--;
                x=x+temp*t;
            }
            printf("%d %d\n",x+temp,y-a/d*(t+1));
        }
        return true;
    }
    int main()
    {
        int a,b;
        while (scanf("%d%d",&a,&b)!=EOF) {
            int x,y;
            if (!linear_equation(a,b,1,x,y))
                printf("sorry\n");
        }
        return 0;
    }