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 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).
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 1, 4, 7,
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 0, 50, 100,
and all the following elements are greater than Vasya‘s favorite integer.
Source
My Solution
被cha了，(┬＿┬)， 原因是用 (b - a) * c >= 0 来表示(b-a) 与 c同号， 或b - a == 0.这里int * int 溢出了，以后还是基本上不要这样，写了。
老老实实的 ((t>=0&&c > 0)|| (t<=0 && c < 0)) 这样写吧。
如果 c == 0, 则 a == b 则YES 否则 NO
如果c != 0, 则看看能不能整除， 同时 (b-a) 要与c同号(如果 b与a不相等)
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
int t = b-a;
if(c == 0){
if(a == b) printf("YES");
else printf("NO");
}
else{
//if(t%c == 0 && (t*c >= 0)) printf("YES"); //!t*c 溢出了
if(t%c == 0 && ((t>=0&&c > 0)|| (t<=0 && c < 0))) printf("YES");
else printf("NO");
}
return 0;
}
Thank you!
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