题目
Description
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.
题解
遍历一下会不会在 x
的时候叫即可
代码
/* By:OhYee Github:OhYee Blog:http://www.oyohyee.com/ Email:oyohyee@oyohyee.com かしこいかわいい? エリーチカ! 要写出来Хорошо的代码哦~ */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <vector> #include <list> #include <queue> #include <stack> #include <map> #include <set> #include <functional> using namespace std; const int maxn = 35; bool Do() { int t,s,x; if(scanf("%d%d%d",&t,&s,&x) == EOF) return false; bool flag = true; for(int i = 0;;i++) { int time = t + i*s; if(x == time) { printf("YES\n"); break; } if(i != 0) { int time = t + i*s + 1; if(x == time) { printf("YES\n"); break; } } if(x < time) { printf("NO\n"); break; } } return true; } int main() { while(Do()); return 0; }