Modular Inverse ZOJ - 3609 （拓展欧几里得）

最新推荐文章于 2019-05-04 07:03:00 发布

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最新推荐文章于 2019-05-04 07:03:00 发布

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分类专栏：数论

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本文介绍了一种通过扩展欧几里得算法来求解模逆元的方法，并提供了一个C++实现示例。针对输入的a和m，输出满足条件ax≡1(mod m)的最小正整数x，若不存在则输出NotExist。

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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

Sample Output

4
Not Exist
8

References

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

题意：求a的逆元x

思路：关于欧几里得的知识看一看这篇博文点击打开链接

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll exgcd(ll a,ll b,ll &x, ll &y){
	if(b==0){
		x=1;
		y=0;
		return a;
	}
	ll r=exgcd(b,a%b,y,x);
	y-=x*(a/b);
	return r;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		ll a,m;
		cin>>a>>m;
		ll x,y;
		ll r=exgcd(a,m,x,y);
		if(r==1){
		    while(x<=0){
		    	x+=m/r;
			}
			cout<<x<<endl;	
		}
		else cout<<"Not Exist"<<endl;
	}
	return 0;
}