codeforces#406 A (拓展欧几里得水题）

A. The Monster

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ....

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 29 19

output

82

input

2 116 12

output

-1

Note

In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.

In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time.

随便找点内容。。

拓展欧几里德, c % gcd(a,b) == 0, 必定有整数解(这里的整数可能是负数)
算法我简单说一下, 具体的可以自己百度 or google

对于
ax + by = gcd(a, b), 由于 gcd(a,b) = gcd(b, a % b) = ...
所以 ax1 + by1 = gcd(a,b) = bx2 + a % b y2 = .... = 1xn + 0yn = gcd(a,b)
至此可以求出xn, 然后反推回去, 就可以得到x1 和 y1 (程序一般直接用递归执行)
于是得到ax1 + by1 = gcd(a,b),那如何得到 ax + by = c的解呢
ax1 + by1 = gcd(a,b) -> a(x1 * c/gcd(a,b)) + b(y1 * c/gcd(a,b)) = c
于是可以求出, 一组 x = x1 * c/gcd(a,b), y = y1 * c/gcd(a,b)
然后你要判断整数解的话, 也是简单, 如果(x,y)是一组解, (x + b/gcd(a,b), y - a/gcd(a,b))也是一组解
(x - b/gcd(a,b), y + a / gcd(a,b))也是一组解
我们可以先求出x是正整数的一组解, 用取模既可
x = (x % b/gcd(a,b) + b/gcd(a,b)) % b/gcd(a,b) //这里保证x >= 0 
y = (c - ax) / b, 这样的x已经是最接近0的数,于是不能在减少了
所以
如果y < 0, 那么无正整数解
如果y > 0有正整数解.

#include<bits/stdc++.h>using namespace std;int gcd(int x,int y){return y==0?x:gcd(y,x%y);}int main(){int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);if((b-d)%(gcd(c,-a))==0){int i=1;while(!((d-b+i*c)%a==0)){i++;}cout<<d+i*c<<endl;}elsecout<<-1<<endl;}