hdu 2669 Romantic 扩展欧几里得

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

思路：exgcd 最后X为最小正解

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if(!b) {
        x=1;y=0;return a;
    }
    ll ans=exgcd(b,a%b,x,y);
    ll temp=x;
    x=y;
    y=temp-a/b*y;
    return ans;
}
int main() {
    ios::sync_with_stdio(false);
    ll a,b,x,y;
    while(cin>>a>>b) {
        ll g=exgcd(a,b,x,y);
        if(g!=1) {
            cout<<"sorry"<<endl;
        } else {
            x=(x%b+b)%b;
            y=(1-a*x)/b;
            cout<<x<<" "<<y<<endl;
        }
    }
    return 0;
}