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=;y=;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!=) {
             cout<<"sorry"<<endl;
         } else {
             x=(x%b+b)%b;
             y=(-a*x)/b;
             cout<<x<<" "<<y<<endl;
         }
     }
     return ;
 }