zoj Modular Inverse 3609 （拓展欧几里德）

ZOJ Problem Set - 3609

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

Each test case contains two integers 0 < a ≤ 1000 and 0 < m

Output

For each test case, output the smallest positive x. If such x

Sample Input

3
3 11
4 12
5 13

Sample Output

4 Not Exist 8

//题意：

给出a,m的值，让你找出一个x使得(a*x)%m==1成立。

拓展欧几里德变形

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
using namespace std;
int gcd(int a,int b,int &x,int &y)
{
	if(!b)
	{
		x=1;
		y=0;
		return a;
	}
	int ans=gcd(b,a%b,x,y);
	int tmp=x;
	x=y;
	y=tmp-a/b*y;
	return ans;
}
int cal(int a,int b,int c)
{
	int x,y,cnt;
	int d=gcd(a,b,x,y);
	if(c%d!=0)
		return -1;
	x*=c/d;
	b/=d;
	if(b<0)
		b-=b;
	cnt=x%b;
	if(cnt<=0)
		cnt+=b;
	return cnt;
}
int main()
{
	int t,num;
	int a,b,c,x,y;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&a,&b);
		num=cal(a,b,1);
		if(num==-1)
			printf("Not Exist\n");
		else
			printf("%d\n",num);
	}
	return 0;
}