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;

}