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ZOJ 3609 Modular Inverse (水题)
Modular Inverse
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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