HDU 2669

扩展欧几里得模板题

ax+by=1

推导： ax+by=gcd(a,b)

          a'x+b'y=gcd(a',b');

   这里a'=b;

  b'=a%b=a-a/b*b;

    那么   bx+(a-a/b*b)y=gcd(a',b')=gcd(a,b)

          ay+b(x-a/b*y)=gcd(a,b)

  由此得出 x'=y  ,y'=x-a/b*y;

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int ext_gcd(int a,int b,int &x,int &y){
    if(b==0){
        x=1;y=0;
        return a;
    }
    int gcd=ext_gcd(b,a%b,x,y);
    int t=x;
    x=y;
    y=t-a/b*y;
    return gcd;
}
void solve(int a,int b){
    int x=0,y=0;
    int g=ext_gcd(a,b,x,y);
    if(g!=1){
        cout<<"sorry"<<endl;
        return ;
    }
    int t=x;
    x=(x%b+b)%b;
    y=y-(x-t)/b*a;
    cout<<x<<' '<<y<<endl;
    return ;
}
int main()
{
    int a,b;
    while(cin>>a>>b){
        solve(a,b);
    }
    return 0;
}