Modular Inverse(zoj3609+欧几里德)

Modular Inverse


Time Limit: 2 Seconds       Memory Limit: 65536 KB

The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1x (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

References


Author: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest

 

 

 

首先我来回顾下欧几里德的几个定理,有助于理解这道题;

   定理一:如果d = gcd(a, b),则必能找到正的或负的整数k和l,使 d = a*x+ b*y。

   定理二:若gcd(a, b) = 1,则方程ax ≡ c (mod b)在[0, b-1]上有唯一解。

   定理三:若gcd(a, b) = d,则方程ax ≡ c (mod b)在[0, b/d - 1]上有唯一解。

 

转载请注明出处:寻找&星空の孩子

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4712

对于ax+by=1;  即ax=1(mod b)      当且仅当gcd(a,b)!=1 的时候,无解!

 

 

 1 #include<stdio.h>

 2 

 3 void exgcd(int a,int b,int &d,int &x,int &y)

 4 {

 5     if(!b){d=a;x=1;y=0;}

 6     else

 7     {

 8         exgcd(b,a%b,d,y,x);

 9         y-=x*(a/b);

10     }

11 }

12 int main()

13 {

14     int T,a,m;

15     scanf("%d",&T);

16     while(T--)

17     {

18         int d,x,y;

19         scanf("%d%d",&a,&m);

20         exgcd(a,m,d,x,y);

21         if(d==1)

22         {

23             while(x<=0)

24             {

25                 x+=m/d;

26             }

27             printf("%d\n",x);

28         }

29         else

30             printf("Not Exist\n");

31     }

32     return 0;

33 }
View Code

 

你可能感兴趣的:(inverse)