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 求ax + by = gcd(a,b)的一组整数解

//    ax + by = gcd(a,b)
//<==>ax + by = b*x1 + (a%b)*y1
//<==>ax + by = b*x1 + (a-(a/b)*b)*y1   //因为a%b = a-(a/b)*b
//<==>ax + by = b*x1 + a*y1 -(a/b)*b*y1
//<==>ax + by = a*y1 + b*(x1-(a/b)*y1)
//得到ｘ,y和之前递归得到的x1,y1之间的关系，即x=y1,y=x1-a/b*y1
//在求gcd(a,b)的时候，每辗转相除一次，就更新一次x,y,临界条件是当a%b == 0时x = 1,y = 0;
#include<bits/stdc++.h>
using namespace std;

int ex_gcd(int a,int b,int &x, int &y)
{
    if(b==0)
    {
        int g=a;
        x=1;
        y=0;
        return g;
    }
    else
    {
        int g =ex_gcd(b,a%b,x,y);
        int x1=y,y1=x-a/b*y;
        x=x1,y=y1;
        return g;
    }
}

int main()
{
    int a,b,x,y,g;
    while(scanf("%d %d",&a, &b) !=EOF)
    {
        g = ex_gcd(a,b,x,y);
        printf("%d*(%d)+%d*(%d)=%d\n",a,x,b,y,g);
        for(int t = 1;t <= 20;t ++)
        {
            printf("%d*(%d)+%d*(%d)=%d\n",a,x+b*t,b,y-a*t,g);//令ｘ = x+b*t,y = y-a*t,ｔ∈Ｚ,可得到此方程的所有整数解
        }
    }
    return 0;
}