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ZOJ 3609 Modular Inverse (水题)

2024-07-18 16:52:34   211人阅读

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-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

33 114 125 13

Sample Output

4Not Exist8

References

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

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121         {printf("1\n");continue;}22         23         int Gcd=gcd(n,m);24         25         if(Gcd>1)26         {printf("Not Exist\n");continue;}27          28         else29             for(int i=1;i<=1000;i++)30                 if((n*i)%m==1)31                 {printf("%d\n",i);break;}   32     }33     return 0;34 }
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