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.
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.
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.
3 10 4
NO
3 10 3
YES
3 8 51
YES
3 8 52
YES
In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4and 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.
思路：用sum表示barks的时间，如果x等于sum的话就输出yes，否则输出no，注意如果x==t的话也是可以的
1 #include<bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 int t,s,x; 6 cin>>t>>s>>x; 7 int sum=0; 8 int i=1; 9 int flag=1; 10 if(t==x) 11 flag=0; 12 else 13 { 14 while(sum<x) 15 { 16 sum=t+i*s; 17 if(sum==x) 18 { 19 flag=0; 20 break; 21 } 22 sum=t+i*s+1; 23 if(sum==x) 24 { 25 flag=0; 26 break; 27 } 28 i++; 29 } 30 } 31 if(flag) 32 cout<<"NO"<<endl; 33 else 34 cout<<"YES"<<endl; 35 return 0; 36 }