zjuoj 3609 Modular Inverse

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3609

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

 

分析：
题目要求给出a和m的值 ， 求出 ax % m == 1 % m成立时的x 的最小值 ， 直接x枚举到m即可。

一开始写的时候没有想到是枚举到m， 后来队友推出m。

 

 

AC代码：

 1 #include<cstdio>

 2 #include<algorithm>

 3 #include<cstring>

 4 #include<queue>

 5 #include<iostream>

 6 #include<stack>

 7 #include<map>

 8 #include<string>

 9 using namespace std;

10 int main(){

11     int n, x, a, m;

12     scanf("%d", &n);

13     while(n--){

14         bool flag = true;

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

16         for(x = 1; x <= m; x++){

17             if((a*x)%m == 1%m){

18                 flag = false;

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

20                 break;

21             }

22         }

23         if(flag){

24             printf("Not Exist\n");

25         }

26     }

27     return 0;

28 }

View Code