Codeforces 697A - Pineapple Incident-优快云博客

本文链接：https://blog.youkuaiyun.com/nare123/article/details/51915623

本文描述了一个有趣的编程问题：一只能够像斗牛犬一样吠叫的松果。它首次吠叫的时间、吠叫间隔以及巴尼计划吃掉松果的时间被设定为变量。任务是判断在巴尼打算吃掉松果的时候，它是否正在吠叫。文章提供了输入输出样例以及一种解决方案。

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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.

$\text{[math]}$

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.

通过写出条件可以得知，所要求的Xn一定属于以下区间的时候是YES

t<=x1<=t+s<=x2<=t+s+1<=x3<=t+2s....

可以得出

0<=x-t-n*s<=s 即当(x-t)%s==0时或0<=x-t<=1是在bark time的

还有一种情况是x<t时是一定不在的，注意特判一下就好。

#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{


    int t,x,s;
    cin>>t>>s>>x;
    if (((x-t)%s==0&&x-t>=0)||x>=t&&(x-t-1)%s==0&&(x-t-1)!=0)
    {
        printf("YES");
    }else
    {
        printf("NO");
    }

}