浙江省省赛J题

     一道数论题，也是道简单题，就是个扩展欧几里得，一直没写过扩展欧几里得的代码，这次还是在朱神的帮助下才写出来的，不过高兴的是1A了，，题目：

Modular Inverse Time Limit: 2 Seconds       Memory Limit: 65536 KB

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

ac代码：

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
int x,y,q;
int gcd(int a,int b){
	if(b==0)
		return a;
	return gcd(b,a%b);
}
int extend_uclid(int a,int b){
	if(b==0){
	  x=1;
	  y=0;
	  q=a;
	  return q;
	}
	else{
	  extend_uclid(b,a%b);
	  int tmp=x;
	  x=y;
	  y=tmp-a/b*y;
	}
}
int main(){
	//freopen("2.txt","r",stdin);
	int numcase;
	int a,m;
	scanf("%d",&numcase);
	while(numcase--){
	  scanf("%d%d",&a,&m);
	  if(m==1){
		  if(a%m==0){
		     puts("1");
		  }
		  else if(m%a==0){
		    printf("%d\n",m/a);
		  }
		  else{
		    puts("Not Exist");
		  }
	  }
	  else{
		int flag=gcd(a,m);
		if(flag!=1)puts("Not Exist");
		else{
	      extend_uclid(a,m);
		  while(x<0){
		    x+=m;
		  }
		  printf("%d\n",x);
		}
	  }
	}
	return 0;
}