hdu-2669 Romantic---扩展欧几里得

Posted 2020-11-05 努力努力再努力x

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=2669

题目大意：

给定a b，求ax+by = 1是否有解，输出x为正数的最小的解

解题思路：

a b互质才有解，下面的方法可求出最小正数的x的解

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll extgcd(ll a, ll b, ll&x, ll&y)//ax+by = gcd(a, b)的解。返回值为gcd
 5 {
 6     ll d = a;
 7     if(b)
 8     {
 9        d = extgcd(b, a % b, y, x);
10        y -= (a / b) * x;
11     }
12     else x = 1, y = 0;
13     return d;
14 }
15 int main()
16 {
17     ll a, b;
18     while(cin >> a >> b)
19     {
20         if(__gcd(a, b) != 1)
21         {
22             cout<<"sorry"<<endl;
23             continue;
24         }
25         ll x, y;
26         extgcd(a, b, x, y);
27         x = ((x % b) + b) % b;
28         y = (1 - a * x) / b;
29         cout<<x<<" "<<y<<endl;
30     }
31     return 0;
32 }