CodeForces - 675A Infinite Sequence （模拟）水

CodeForces - 675A

Infinite Sequence

Time Limit:                                                        1000MS                        Memory Limit:                                                        262144KB                        64bit IO Format:                            %I64d & %I64u                       

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Description

Vasya likes everything infinite. Now he is studying the properties of a sequences, such that its first element is equal toa (s₁ = a), and the difference between any two neighbouring elements is equal toc (s_i - s_i - 1 = c). In particular, Vasya wonders if his favourite integer b appears in this sequence, that is, there exists a positive integeri, such that s_i = 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

Sample Output

Hint

Source

Codeforces Round #353 (Div. 2)

//题意：输入a,b,c；

a表示序列的第一个元素的值s1，b是要找的元素的值，c是s(i)-s(i-1)的值，问b是否在这个序列中？

//思路：

直接模拟就行了

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