Codeforces Round #353 (Div. 2)-A. Infinite Sequence(模拟)

A. Infinite Sequence
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 integer bappears 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 ab 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).

Examples
input
1 7 3
output
YES
input
10 10 0
output
YES
input
1 -4 5
output
NO
input
0 60 50
output
NO
Note

In the first sample, the sequence starts from integers 147, so 7 is its element.

In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.

In the third sample all elements of the sequence are greater than Vasya's favorite integer.

In the fourth sample, the sequence starts from 050100, and all the following elements are greater than Vasya's favorite integer.



题意:给出s1=a,si=b,找等差数列是否符合c。坑点在i=1时,任何c都是符合的。

AC代码:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
#define QWQ ios::sync_with_stdio(0)
typedef unsigned __int64 LL;
typedef  __int64 ll;
const int T = 200000+50;
const int mod = 1000000007;
const double PI = 3.1415926535898;

int main()
{
#ifdef zsc
	freopen("input.txt","r",stdin);
#endif

	ll a,b,c;
	while(~scanf("%I64d%I64d%I64d",&a,&b,&c))
	{
		if(a==b){
			printf("YES\n");
		}
		else if(c==0){
			if(b==a)printf("YES\n");
			else printf("NO\n");
		}
		else if(c>0){
			if(a>=b)printf("NO\n");
			else {
				if((b-a)%c==0)printf("YES\n");
				else printf("NO\n");
			}
		}
		else {
			if(a<=b)printf("NO\n");
			else {
				if((a-b)%(-c)==0)printf("YES\n");
				else printf("NO\n");
			}
		}
	}

	return 0;
}


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