Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) A
Arpa is researching the Mexican wave.
There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.
- At time 1, the first spectator stands.
- At time 2, the second spectator stands.
- ...
- At time k, the k-th spectator stands.
- At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
- At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
- ...
- At time n, the n-th spectator stands and the (n - k)-th spectator sits.
- At time n + 1, the (n + 1 - k)-th spectator sits.
- ...
- At time n + k, the n-th spectator sits.
Arpa wants to know how many spectators are standing at time t.
The first line contains three integers n, k, t (1 ≤ n ≤ 109, 1 ≤ k ≤ n, 1 ≤ t < n + k).
Print single integer: how many spectators are standing at time t.
10 5 3
3
10 5 7
5
10 5 12
3
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At t = 0 ---------- number of standing spectators = 0.
- At t = 1 ^--------- number of standing spectators = 1.
- At t = 2 ^^-------- number of standing spectators = 2.
- At t = 3 ^^^------- number of standing spectators = 3.
- At t = 4 ^^^^------ number of standing spectators = 4.
- At t = 5 ^^^^^----- number of standing spectators = 5.
- At t = 6 -^^^^^---- number of standing spectators = 5.
- At t = 7 --^^^^^--- number of standing spectators = 5.
- At t = 8 ---^^^^^-- number of standing spectators = 5.
- At t = 9 ----^^^^^- number of standing spectators = 5.
- At t = 10 -----^^^^^ number of standing spectators = 5.
- At t = 11 ------^^^^ number of standing spectators = 4.
- At t = 12 -------^^^ number of standing spectators = 3.
- At t = 13 --------^^ number of standing spectators = 2.
- At t = 14 ---------^ number of standing spectators = 1.
- At t = 15 ---------- number of standing spectators = 0.
题意:自己读一读就行
解法:暗中观察
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,k,t;
cin>>n>>k>>t;
if(k>=t){
cout<<t<<endl;
}else if(t<=n){
cout<<k<<endl;
}else if(n+k<=t){
cout<<""<<endl;
}else{
cout<<k-(t-n)<<endl;
}
return ;
}
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