ZOJ 3609 Modular Inverse (水题)

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

References

• http://en.wikipedia.org/wiki/Modular_inverse

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

 

简单来说就是要求给定n,m 求一个x使得 (n*x)%m=1, 如果x存在输出最小正整数x，否则输出Not Exist

注意m=1的情况，因为任何数对1取模会等于0，但是这里要求输出最小正整数，所以输出1

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<stdlib.h>

 4 #include<algorithm>

 5 using namespace std;

 6 int gcd(int a,int b)

 7 {

 8     return b?gcd(b,a%b):a;

 9 }

10 int main()

11 {

12     //freopen("in.txt","r",stdin);

13     int kase;

14     scanf("%d",&kase);

15     while(kase--)

16     {

17         int n,m;

18         scanf("%d %d",&n,&m);

19         

20         if(m==1)//当m=1时，数字对1取模等于0，存在这个数字，但是这里要输出最小的正整数，所以输出1

21         {printf("1\n");continue;}

22         

23         int Gcd=gcd(n,m);

24         

25         if(Gcd>1)

26         {printf("Not Exist\n");continue;}

27          

28         else

29             for(int i=1;i<=1000;i++)

30                 if((n*i)%m==1)

31                 {printf("%d\n",i);break;}   

32     }

33     return 0;

34 }

View Code