CodeForces675AInfinite Sequence

Description

Vasya likes everything infinite. Now he is studying the properties of a sequence s, such that its first element is equal to a (s1 = a),
and the difference between any two neighbouring elements is equal to c (si - si - 1 = c).
In particular, Vasya wonders if his favourite integerb appears in this sequence, that is, there exists a positive integer i, such
that si = b. Of course, you are the person he asks for a help.

Input

The first line of the input contain three integers a, b and c ( - 109 ≤ a, b, c ≤ 109) —
the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.

Output

If b appears in the sequence s print "YES" (without quotes), otherwise print "NO"
(without quotes).

Sample Input

Input

1 7 3

Output

YES

Input

10 10 0

Output

YES

Input

1 -4 5

Output

NO

Input

0 60 50

Output

NO

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int main()
{
int a,b,c;
while(scanf("%d %d %d",&a,&b,&c)!=EOF)
{
if(c!=0)
{
if(b>a)
{
if(c<0)
printf("NO\n");
else
{
if((b-a)%c==0)
printf("YES\n");
else
printf("NO\n");
}
}
else if(b<a)
{
if(c>0)
printf("NO\n");
else
{
if((b-a)%c==0)
printf("YES\n");
else
printf("NO\n");
}
}
else if(a==b)
printf("YES\n");
}
else
{
if(a==b)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}

题意：给你最初值a和公差c，再给你一个b，问b是否在这个数列中。
思路：分成几种情况判断。例如a>b时如果c<=0那么就不可能有输出NO，细心些考虑所有的情况就能ac啦。注意不能直接对a+c一直下去会超时，直接用b-a对c取余，如果值为0就可以。但注意判断符号和彼此大小值。例如a=3，b=5，c=-2.虽然（b-a）%c==0，但是并不成立输出NO。