A - Typhoon Damage Survey
【题目来源】
AtCoder：A - Typhoon Damage Survey
【题目描述】
Takahashi is the caretaker of an orchard. There are \(N\) fruit trees planted in this orchard, numbered \(1, 2, \ldots, N\). The height of fruit tree \(i\) is \(H_i\).
高桥是一个果园的看护人。这个果园中种有 \(N\) 棵果树，编号为 \(1, 2, …, N\)。果树 \(i\) 的高度为 \(H_i\)。
Last night, a large typhoon passed through this area. Due to the typhoon's strong winds, fruit trees with height \(T\) or less had not yet developed sufficient roots and all fell over. That is, fruit tree \(i\) has fallen if and only if \(H_i \leq T\).
昨晚，一场强台风经过了这个地区。由于台风的强风，高度不超过 \(T\) 的果树因根系尚未充分发育而全部倒伏。也就是说，果树 \(i\) 倒伏当且仅当 \(H_i ≤ T\)。
To restore the orchard, Takahashi decided to replant all the fallen fruit trees. The cost to replant fruit tree \(i\) and restore it to its original state is \(C_i\).
为了恢复果园，高桥决定重新种植所有倒伏的果树。重新种植果树 \(i\) 并将其恢复到原始状态的成本是 \(C_i\)。
Find the total cost required to restore all fallen fruit trees to their original state.
求将所有倒伏的果树恢复到原始状态所需的总成本。
【输入】
Input is given from standard input in the following format.
\(N\) \(T\)
\(H_1\) \(H_2\) \(\ldots\) \(H_N\)
\(C_1\) \(C_2\) \(\ldots\) \(C_N\)
- The first line contains an integer \(N\) representing the number of fruit trees and an integer \(T\) representing the height threshold for trees falling, separated by a space.
- The second line contains integers \(H_1, H_2, \ldots, H_N\) representing the height of each fruit tree, separated by spaces.
- The third line contains integers \(C_1, C_2, \ldots, C_N\) representing the restoration cost for replanting each fruit tree, separated by spaces.
【输出】
Output in a single line the total cost required to restore all fallen fruit trees (fruit trees \(i\) satisfying \(H_i \leq T\)). If no fruit trees have fallen, output \(0\). Include a newline at the end.
【输入样例】
5 3
1 5 3 4 2
10 20 30 40 50
【输出样例】
90
【解题思路】
【代码详解】
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 200005;int n, t, h[N], c[N], ans; // n: 物品数量，t: 阈值// h: 高度数组，c: 价值数组，ans: 总价值signed main()
{cin >> n >> t; // 读入物品数量和阈值for (int i = 1; i <= n; i++){cin >> h[i]; // 读入每个物品的高度}for (int i = 1; i <= n; i++){cin >> c[i]; // 读入每个物品的价值}// 统计高度不超过阈值的物品的总价值for (int i = 1; i <= n; i++){if (h[i] <= t) // 如果高度不超过阈值{ans += c[i]; // 累加价值}}cout << ans << endl; // 输出总价值return 0;
}
【运行结果】
5 3
1 5 3 4 2
10 20 30 40 50
90