Modular Inverse(扩展欧几里得)

Posted 2022-06-27 qq-1585047819

Modular Inverse

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1≡x (mod m). This is equivalent to ax≡1 (mod m).

Input

There are multiple test cases. The first line of input is an integer T ≈ 2000 indicating the number of test cases.

Each test case contains two integers 0 < a ≤ 1000 and 0 < m ≤ 1000.

Output

For each test case, output the smallest positive x. If such x doesn‘t exist, output "Not Exist".

Sample Input

3
3 11
4 12
5 13

Sample Output

4
Not Exist
8

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <map>
 5 using namespace std;
 6 
 7 typedef long long ll;
 8 
 9 int t;
10 int a,b,X,Y;
11 
12 int gdc(int a,int b)
13     return b==0?a:gdc(b,a%b);
14 
15 
16 void exgdc(int a,int b,int &X,int &Y)
17     if(b==0)
18         X=1;
19         Y=0;
20         return;
21     
22     exgdc(b,a%b,X,Y);
23     int temp=X;
24     X=Y;
25     Y=temp-a/b*Y;
26 
27 
28 
29 int main()
30     ios::sync_with_stdio(false);
31     while(cin>>t)
32         while(t--)
33             cin>>a>>b;
34             int GDC=gdc(a,b);
35             if(1%GDC!=0) cout << "Not Exist" << endl;continue;
36             exgdc(a,b,X,Y);
37             int res=X*(1/GDC);
38             int ans=b/GDC;
39             if(ans<0) ans=-ans;
40             res%=ans;
41             if(res<=0) res+=ans;   //题目说的是正数
42             cout << res << endl;
43         
44     
45     return 0;
46 

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