A strip of N cells must each be painted one of 3 c...
Tạo vào: 26 tháng 4, 2026
Trả lời bằng GPT-5.5 Thinking bởi Chat01
Câu hỏi
A strip of N cells must each be painted one of 3 colors (R, G, B
Painting cell i with color c costs cost didanis
[c]
A streak of color c is a maximal run of consecutive cells all painted c. A streak of length Learns a streak bonus of Lx Q (Q is a given constant)
Minimize total painting cost-total streak bonuses
Find the minimum total cost.
Input Format
The first line contains a integer, n, denoting the number of cells.
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The second line contains a integer, q, denoting the bonus per cell in any same-color streak.
Each of the n lines contains 3 space-separated integers, representing row i of cost.
alia.
Constraints
1 <= n <= 10^5
0 <= q <= 10^4
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1 <= cost[i][j] <=
10^4
Sample Test Cases
Case 1
Input:
3
10
5 20 30
50 5 40
100 100 5
alidanish0666
Output:
-15
Explanation:
n = 3 q = 10 Cheapest per cell: R(5), G(5), B(5) paint cost 3 * 10 = 30 Net = 15 alidanish06625 boxed 3theta = - 15 . . Bonus =
Case 2
Input:
2
0
10 20 30
30 20 10
Output:
20
alidanish0666
Explanation:
n = 2 q = 0 Cheapest per cell: R(10), B(10) paint cost 20. Bonus = 2 x 0 0. Net 200 = 20.
Case 3
Input:
4
3
8 15 6
10 4 20
7 12 9
3 18 11
alidanish0666
Output:
8
Explanation:
pish0666
Total paint cost Streak bonusev +4+7+ 3 = 20 every cell contributes exactly Q to the total bonus regardless of coloring, so total bonus =NxQ= 4 x 3 12 Net 2012 = 8
#include <bits/stdc++.h>
using namespace std;
int solve(int n, int q, vector<vector<int>>& cost) ( } // Write your solution here
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n; cin >> n;
int q; cin >> q;
vector<vector<int>> cost(n, vector<int>(3));
for (int i = 0; i < n; i++)
for (int j = 0; j < 3; j++) cin >> cost[i][j];
auto result = solve(n, q, cost);
cout << result << endl;
return 0;
}
Please Give me proper code without explanation also I mentioned in the last
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Trả lời
cpp#include <bits/stdc++.h> using namespace std; long long solve(int n, int q, vector<vector<int>>& cost) { long long totalCost = 0; for (int i = 0; i < n; i++) { totalCost += min({cost[i][0], cost[i][1], cost[i][2]}); } long long totalBonus = 1LL * n * q; return totalCost - totalBonus; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int n; cin >> n; int q; cin >> q; vector<vector<int>> cost(n, vector<int>(3)); for (int i = 0; i < n; i++) { for (int j = 0; j < 3; j++) { cin >> cost[i][j]; } } cout << solve(n, q, cost) << endl; return 0; }