hdu 2669 Romantic 扩展欧几里得

Posted 2020-10-14 lemonsbiscuit

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：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll exgcd(ll a, ll b, ll &x, ll &y) {
 5     if(!b) {
 6         x=1;y=0;return a;
 7     }
 8     ll ans=exgcd(b,a%b,x,y);
 9     ll temp=x;
10     x=y;
11     y=temp-a/b*y;
12     return ans;
13 }
14 int main() {
15     ios::sync_with_stdio(false);
16     ll a,b,x,y;
17     while(cin>>a>>b) {
18         ll g=exgcd(a,b,x,y);
19         if(g!=1) {
20             cout<<"sorry"<<endl;
21         } else {
22             x=(x%b+b)%b;
23             y=(1-a*x)/b;
24             cout<<x<<" "<<y<<endl;
25         }
26     }
27     return 0;
28 }

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