扩欧应用

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.

题目大意：

有一个怪物在追赶  Rick 和 Morty    他们两个人很害怕 所以不停的尖叫    

Rick在  b, b + a, b + 2a, b + 3a, ...   时刻尖叫   

Morty在   d, d + c, d + 2c, d + 3c, .... 时刻尖叫

问两个人会不会在同一时刻尖叫

分析：

扩展欧几里得  略微分析便可得到方程

若想两人同时尖叫   需满足   b + a*x = d + c*y   --->    a*x - c*y = d - b

求出 满足方程的最小非负整数对   x，y

AC代码：

[cpp] view plain copy

print?

1. #include<stdio.h>  
2. #include <math.h>  
3. int gcd(int a,int b,int &x,int &y){  
4.     if (b==0){  
5.         x=1,y=0;  
6.         return a;  
7.     }  
8.     int ans=gcd(b,a%b,y,x);  
9.     y-=(a/b)*x;  
10.     return ans;  
11. }  
12. int main (){  
13.     int a,b,c,d;  
14.     while (scanf ("%d%d%d%d",&a,&b,&c,&d)!=EOF){  
15.         int x,y;  
16.         int g=gcd(a,-c,x,y);  
17.         if ((d-b)%g!=0){//    不满足条件   
18.             printf ("-1\n");  
19.             continue;  
20.         }  
21.         x=x*(d-b)/g;//     求出方程的x   
22.         //        接下来三行求出最小非负整数x   
23.         x=x%(-c/g);   
24.         if (x<0)  
25.             x+=fabs(-c/g);   
26.         y=((d-b)-a*x)/(-c); //     求出方程的y   
27.         while (y<0){  
28.             x=x+fabs(-c/g);//          
29.             y=((d-b)-a*x)/(-c);  
30.         }  
31.         printf ("%d\n",b+x*a);    
32.     }  
33.     return 0;  
34. }  

#include<stdio.h>#include <math.h>int gcd(int a,int b,int &x,int &y){if (b==0){x=1,y=0;return a;}int ans=gcd(b,a%b,y,x);y-=(a/b)*x;return ans;}int main (){int a,b,c,d;while (scanf ("%d%d%d%d",&a,&b,&c,&d)!=EOF){int x,y;int g=gcd(a,-c,x,y);if ((d-b)%g!=0){//    不满足条件 printf ("-1\n");continue;}x=x*(d-b)/g;//     求出方程的x //        接下来三行求出最小非负整数x x=x%(-c/g); if (x<0)x+=fabs(-c/g); y=((d-b)-a*x)/(-c);//     求出方程的y while (y<0){x=x+fabs(-c/g);//        y=((d-b)-a*x)/(-c);}printf ("%d\n",b+x*a);}return 0;}

其实这题可以水

AC代码：

[cpp] view plain copy

print?

1. #include<stdio.h>  
2. int main (){  
3.     int a,b,c,d;  
4.     scanf("%d%d%d%d",&a,&b,&c,&d);  
5.     int flag=1;  
6.     for (int i=1;i<=200&&flag;i++){  
7.         for (int j=1;j<=200&&flag;j++){  
8.             if ((i*a-j*c)==(d-b+a-c)){  
9.                 printf ("%d\n",d+(j-1)*c);  
10.                 flag=0;  
11.             }  
12.         }     
13.     }   
14.     if (flag)  
15.         printf ("-1\n");  
16.     return 0;  
17. }