Modular Inverse （拓展欧几里得求逆元）

The modular modular multiplicative inverse of an integer a modulo m is an integer xsuch 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

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

今晚实在是想吐槽一下这个题，很无奈啊，本来想四题签到跑路，被这个题卡了2个小时，首先暴力枚举得没过，然后翻板子在拓展欧几里的找到了类似得题目，把板子搞上，疯狂wa，wa到自闭，结果是一组m为1时的特例为1，程序输出为0，还是太菜了，真************(疯狂喷自己）

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int  extend_gcd(int a,int b,int &x,int &y)
{
	if(a==0&&b==0)
	{
		return -1;
	}
	if(b==0)
	{
		x=1;y=0;
		return a;
	}
	int  d=extend_gcd(b,a%b,y,x);
	y-=a/b*x;
	return d;
}
int mod_reverse(int a,int  n)
{
	int x,y;
	int d=extend_gcd(a,n,x,y);
	if(d==1)return(x%n+n)%n;
	else return -1;
}
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int  a,n;
		cin>>a>>n;
		if(n==1)
		{
			cout<<1<<endl;
			continue;
		}
		if(__gcd(a,n)!=1)
		{
			cout<<"Not Exist"<<endl;
			continue;
		}
		cout<<mod_reverse(a,n)<<endl;
	}

	return 0;
}