1 solutions
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1
#include <bits/stdc++.h> using namespace std; int n, k, a[100005], sum; int main() { cin >> n >> k; for (int i = 1;i <= n;i++) { cin >> a[i]; } for (int i = 1;i <= min(10, k);i++) { for (int j = 1;j <= n;j++) { a[j] = ceil(sqrt(a[j]) * 10); } } for (int i = 1;i <= n;i++) { sum += a[i]; } double x = sum, y = n; cout << fixed << setprecision(2) << x / y; return 0; }
- 1
Information
- ID
- 1084
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 8
- Tags
- (None)
- # Submissions
- 76
- Accepted
- 13
- Uploaded By