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).
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.
For each test case, output the smallest positive x. If such x doesn‘t exist, output "Not Exist".
3 3 11 4 12 5 13
4 Not Exist 8
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 }
ZOJ 3609 Modular Inverse (水题),布布扣,bubuko.com
原文：http://www.cnblogs.com/clliff/p/3926582.html
