Codeforces Round #362 (Div. 2) A. Pineapple Incident

A. Pineapple Incident
time limit per test

1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in seconds) it barks for the first time. Then every s seconds after it, it barks twice with 1 second interval. Thus it barks at times t, t + s, t + s + 1, t + 2s, t + 2s + 1, etc.

Barney woke up in the morning and wants to eat the pineapple, but he can’t eat it when it’s barking. Barney plans to eat it at time x (in seconds), so he asked you to tell him if it’s gonna bark at that time.
Input

The first and only line of input contains three integers t, s and x (0 ≤ t, x ≤ 109, 2 ≤ s ≤ 109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.
Output

Print a single “YES” (without quotes) if the pineapple will bark at time x or a single “NO” (without quotes) otherwise in the only line of output.
Examples
Input

3 10 4

Output

NO

Input

3 10 3

Output

YES

Input

3 8 51

Output

YES

Input

3 8 52

Output

YES

Note

In the first and the second sample cases pineapple will bark at moments 3, 13, 14, …, so it won’t bark at the moment 4 and will bark at the moment 3.

In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, …, so it will bark at both moments 51 and 52.

题意：如题，判断t能不能到达s，每次+x，要判断当前和当前+1

思路：直接模拟就可以了

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headint main(){    int t,x,s;scanf("%d%d%d",&t,&x,&s);    int bz=0;    if(t==s) bz=1;t+=x;    for(int i=t;i<=s;i+=x){        if(i==s||i+1==s){            bz=1;            break;        }    }    printf(bz?"YES\n":"NO\n");    return 0;}