ex_gcd求不定方程的最小正整数解

#include<bits/stdc++.h>
using namespace std;
int gcd(int a,int b) {return b?gcd(b,a%b):a;}
int exgcd(int &x,int &y,int a,int b)
{
    if(!b)
    {
        x=1;
        y=0;
        return a;
    }
    int r=exgcd(x,y,b,a%b);
    int t=x;
    x=y;
    y=t-a/b*y;
    return r;
}
bool cal(int &x,int &y,int a,int b,int c)
{
    int d=exgcd(x,y,a,b);
    if(c%d) return false;
    int k=c/d;
    x*=k;
    y*=k;
    return true;
}
int main()
{
    int x,y,a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    if(cal(x,y,a,b,c))
    {
        printf("%d %d\n",x,y);//输出任意解 
        int tx=x;
        x%=b;
        if(x<=0) x+=(int)abs(b*1.0);
        int k=(tx-x)/b;
        y += k*a;
        printf("%d %d\n",x,y);//输出最小正整数解 
    }
    else puts("No answer!");//无解
    return 0;
}