Modular Inverse(zoj3609+欧几里德)
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
3 3 11 4 12 5 13
Sample Output
4 Not Exist 8
References
Author: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest
首先我来回顾下欧几里德的几个定理,有助于理解这道题;
定理一:如果d = gcd(a, b),则必能找到正的或负的整数k和l,使 d = a*x+ b*y。
定理二:若gcd(a, b) = 1,则方程ax ≡ c (mod b)在[0, b-1]上有唯一解。
定理三:若gcd(a, b) = d,则方程ax ≡ c (mod b)在[0, b/d - 1]上有唯一解。
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题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4712
对于ax+by=1; 即ax=1(mod b) 当且仅当gcd(a,b)!=1 的时候,无解!
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 3 void exgcd(int a,int b,int &d,int &x,int &y) 4 { 5 if(!b){d=a;x=1;y=0;} 6 else 7 { 8 exgcd(b,a%b,d,y,x); 9 y-=x*(a/b); 10 } 11 } 12 int main() 13 { 14 int T,a,m; 15 scanf("%d",&T); 16 while(T--) 17 { 18 int d,x,y; 19 scanf("%d%d",&a,&m); 20 exgcd(a,m,d,x,y); 21 if(d==1) 22 { 23 while(x<=0) 24 { 25 x+=m/d; 26 } 27 printf("%d\n",x); 28 } 29 else 30 printf("Not Exist\n"); 31 } 32 return 0; 33 }
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