Codeforces 898 B（拓展欧几里得）

 1 #include<iostream>
 2 using namespace std;
 3 #define ll long long
 4 ll gcd(ll a, ll b)
 5 {
 6     return b? gcd(b, a%b) : a;
 7 }
 8 
 9 void ex_gcd(ll a, ll b, ll &x, ll &y, ll &d)
10 {
11     if(!b)
12     {
13         x = 1, y = 0, d = a;
14         return ;
15     }
16     ex_gcd(b, a%b, y, x, d);
17     y -= a/b * x;
18     return ;
19 }
20 int main()
21 {
22     ll m, a, b, x, y;
23     cin >> m >> a >> b;
24     ll d = gcd(a, b);
25     if(m%d)
26     {
27         cout << "NO\n";
28         return 0;
29     }
30     ex_gcd(a, b, x, y, d);
31     x *= m/d, y *= m/d;
32     a /= d, b /= d;
33     if(x<0 && y <0)
34     {
35         cout << "NO\n";
36     }
37     else if(x < 0)
38     {
39         x = -x;
40         ll z;
41         if(x%b) z = x/b+1; //找到最小的正数x
42         else z = x/b;
43         x = -x + z * b;
44         y = y  -  z*a;
45         if(y < 0)
46             cout << "NO\n";
47         else cout << "YES\n" << x << ‘ ‘ << y << endl;
48     }
49     else if(y < 0)
50     {
51         y = -y;
52         ll z;
53         if(y%a) z = y/a+1;
54         else z = y/a;
55         y = -y + z*a;
56         x = x - z*b;
57         if(x < 0)
58             cout << "NO\n";
59         else cout << "YES\n" << x << ‘ ‘ << y << endl;
60     }
61     else
62         cout << "YES\n" << x << ‘ ‘ << y << endl;
63     return 0;
64 }