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C++ code to number of standing spectators at time t
Suppose we have three numbers n, k and t. Amal is analyzing Mexican waves. There are n spectators numbered from 1 to n. They start from time 0. At time 1, first spectator stands, at time 2, second spectator stands. At time k, kth spectator stands then at time (k+1) the (k+1) th spectator stands and first spectator sits, at (k+2), the (k+2)th spectator stands but 2nd one sits, now at nth time, nth spectator stands and (n-k)th spectator sits. At time (n+1), the (n+1-k)th spectator sits and so on. We have to find the number of spectators stands at time t.
So, if the input is like n = 10; k = 5; t = 3, then the output will be 3, because before 5, no one will sit so all spectators from 1 to 3 are standing.
Steps
To solve this, we will follow these steps −
return minimum of t, k and (n + k - t)
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
int solve(int n, int k, int t){
return min({ t, k, n + k - t });
}
int main(){
int n = 10;
int k = 5;
int t = 3;
cout << solve(n, k, t) << endl;
}Input
10, 5, 3
Output
3
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