codeforce 787 A. The Monster

最新推荐文章于 2022-05-26 19:50:53 发布

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本文解析了一道关于两个角色在被怪物追逐过程中尖叫同步问题的算法题。通过使用扩展欧几里得算法来寻找两个尖叫序列首次同步的时间点，详细介绍了如何在满足特定条件下求解此类问题的方法。

A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times b, b + a, b + 2a, b + 3a, ... and Morty screams at times d, d + c, d + 2c, d + 3c, ....

$\text{[math]}$

The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time.

Input

The first line of input contains two integers a and b (1 ≤ a, b ≤ 100).

The second line contains two integers c and d (1 ≤ c, d ≤ 100).

Output

Print the first time Rick and Morty will scream at the same time, or - 1 if they will never scream at the same time.

Examples

input
20 2
9 19

output
82

input
2 1
16 12

output
-1

这个题目使我对于扩展欧几里得的理解更深了。

b+x*a=d+y*c -->a*x-c*y=d-b 因为适用条件是a、c为不完全为零的非负整数。

所以这个题目潜在的要求是：在多组解中选出x>0，y<0的情况。

具体细节详见代码注释；

#include<stdio.h>
int ans;
int Exgcd(int a,int b,int &x,int &y)
{
	if(b==0)
	{
		x=1; y=0;
		return a;
	}
	int gg=Exgcd(b,a%b,y,x);
	y-=a/b*x;
	return gg;
}

int cal(int a,int b,int c,int d)
{
	int x,y;
	int gg=Exgcd(a,c,x,y);
	if((d-b)%gg!=0)
		return 0;
	else
	{
		int m=c/gg;
		printf("xy = %d%d",x,y);
		x=(d-b)/gg*x;	//先按照gg与d-b倍数关系转化 
		y=(d-b)/gg*y;
		x=((x%m)+m)%m;	//将x转化为最小正整数 
		y=(d-b-a*x)/c;
		while(y>0)	//在y大于0的时候不断增大x。 
		{
			x+=m;	//在通解中找出特解。 
			y=(d-b-a*x)/c;
		}
		ans=b+a*x;	
	}
	return 1;
}

int main()
{
	int a,b,c,d;
	while(~scanf("%d%d%d%d",&a,&b,&c,&d))
	{
		if(cal(a,b,c,d))
			printf("%d\n",ans);
		else
			printf("-1\n");
	}
	return 0;
}