codeforces 697A Pineapple Incident_a. pineapple incident time limit per test1 second -优快云博客

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本文探讨了一个趣味算法问题：一个特殊能吠叫的‘松果’遵循特定的时间间隔进行吠叫。根据首次吠叫时间、吠叫间隔及巴尼打算吃掉松果的时间，判断该时刻松果是否会吠叫。通过分析给出的样例输入输出，可以快速掌握问题解决技巧。

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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 
    
      Copy
    
3 10 4

     output 
    
      Copy
    
NO

     input 
    
      Copy
    
3 10 3

     output 
    
      Copy
    
YES

     input 
    
      Copy
    
3 8 51

     output 
    
      Copy
    
YES

     input 
    
      Copy
    
3 8 52

     output 
    
      Copy
    
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.

水题，注意t,t+s,t+s+1，这三个要单独判断，后边的就是规律了

#include<bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
const int inf = 0x3f3f3f3f;  
int main()   
{  
    // freopen("shuju.txt","r",stdin);
    int t,s,x;
    cin>>t>>s>>x;
    int flag=1;
    if(x==t||x==t+s||x==t+s+1)
    {
        printf("YES\n");
        flag=0;
    }
    else if(x<t+s+1)
    {
        printf("NO\n");
        flag=0;
    }
    if(flag)
    {
        if((x-t)%s==0||(x-1-t)%s==0)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;  
}