Infinite 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 integer b 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
正确代码：

#include<iostream>using namespace std;int main(void){    int a, b, c;    cin>>a>>b>>c;    if(c == 0)    {        if(a == b)        cout<<"YES";        else        cout<<"NO";    }    else    {        if(((b-a)%c == 0) && (((b-a)>= 0 && c > 0)||((b-a)<= 0 && c < 0)))        cout<<"YES";        else        cout<<"NO";    }    return 0;}

以下为错误代码段：

#include<stdio.h>#include<iostream>using namespace std;int main(void){    long long a, b, c;    cin>>a>>b>>c;    if(c == 0 && a == b)    {        printf("YES\n");    }    else if(c == 0 && a != b)    {        printf("NO\n");    }    else    {        long long t;        t = (b-a)%c;        //cout<<t;        if(t == 0)        {            if((a > b && c < 0) || (a < b && c > 0 ))            {                printf("YES");            }            else            {                printf("NO");            }        }        else        {            printf("NO");        }    }    return 0;}

个人认为两段代码，逻辑无差别，但是第二个就不对，，，请个位路过的朋友帮忙看看吧~