暑期学习日记二—利用扩展欧几里得求逆元

 1 int exgcd(int a, int b, int &x, int &y)
 2 {
 3     if (b == 0)
 4     {
 5         x = 1;
 6         y = 0;
 7         return a;
 8     }
 9 
10     int mod = exgcd(b, a%b, x, y);
11     int tmp = x;
12     x = y;
13     y = tmp - (a / b)*y;
14     return mod;
15 }

1 int Inverse_element(int a, int n)
2 {
3     int gcd, x, y;
4     gcd = exgcd(a, n, x, y);
5     cout << x << endl;
6     if (gcd == 1)return (x%n + n) % n;
7     else return -1;
8 }

 1 #include<iostream>
 2 #include<malloc.h>
 3 #include<string>
 4 #include<cmath>
 5 #include<conio.h>
 6 
 7 using namespace std;
 8 
 9 int exgcd(int a, int b, int &x, int &y)
10 {
11     if (b == 0)
12     {
13         x = 1;
14         y = 0;
15         return a;
16     }
17 
18     int mod = exgcd(b, a%b, x, y);
19     int tmp = x;
20     x = y;
21     y = tmp - (a / b)*y;
22     return mod;
23 }
24 
25 int Inverse_element(int a, int n)
26 {
27     int gcd, x, y;
28     gcd = exgcd(a, n, x, y);
29     cout << x << endl;
30     if (gcd == 1)return (x%n + n) % n;
31     else return -1;
32 }
33 
34 int main(int argc, const char * argv[])
35 {
36     int a, n;
37     cin >> a >> n;
38     int mod = Inverse_element(a, n);
39     cout << mod <<endl;
40     _getch();
41     return 0;
42 }