扩展欧几里得算法模板题 zoj 3609

Modular Inverse

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

简单模板：

/** ************************************
//考察知识点：扩展欧几里得算法模板；
求乘法逆元的代码段：
int cal(int a,int b,int c)
{
	int x,y;
	int gcd=e_gcd(a,b,x,y);
	if(c%gcd!=0)
	return -1;
	x*=c/gcd;
	b=abs(b);
	int ans=x%b;
	if(ans<=0)
	ans+=b;
	return ans;
}
扩展欧几里德算法模板：
int e_gcd(int a,int b,int &x,int &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return a;
	}
	int ans=e_gcd(b,a%b,x,y);
	int temp=x;
	x=y;
	y=temp-a/b*y;
	return ans;
}
***************************************/
#include<stdio.h>
#include<stdlib.h>
int e_gcd(int a,int b,int &x,int &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return a;
	}
	int ans=e_gcd(b,a%b,x,y);
	int temp=x;
	x=y;
	y=temp-a/b*y;
	return ans;
}
int cal(int a,int b,int c)
{
	int x,y;
	int gcd=e_gcd(a,b,x,y);
	if(c%gcd!=0)
	return -1;
	x*=c/gcd;
	b=abs(b);
	int ans=x%b;
	if(ans<=0)
	ans+=b;
	return ans;
}
int main()
{
	int t,a,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&a,&m);
		int ans=cal(a,m,1);
		if(ans==-1)
		printf("Not Exist\n");
		else
		printf("%d\n",ans);
	}
	return 0;
}