A - Battery Level and Charger Arrival
【题目来源】
AtCoder：A - Battery Level and Charger Arrival (atcoder.jp)
【题目描述】
Takahashi manages \(N\) smartphones. The current battery level of the \(i\)-th smartphone is \(W_i\) percent.
高桥管理着 \(N\) 部智能手机。第 \(i\) 部手机的当前电量为 \(W_i\) 百分比。
The battery of a smartphone decreases by \(D\) percentage points per day if not charged.
如果不充电，手机的电池电量每天下降 \(D\) 个百分点。
Takahashi is considering ordering from an online charger store. If he places an order, the charger will arrive in exactly \(K\) days. Since Takahashi currently has no charger, he cannot charge any smartphone during the \(K\) days until it arrives.
高桥正在考虑从一家在线充电器商店下单。如果他下单，充电器将在恰好 \(K\) 天后送达。由于高桥目前没有充电器，在充电器送达前的 \(K\) 天内，他无法为任何手机充电。
After \(K\) days, when the charger arrives, the battery level of the \(i\)-th smartphone is considered to be \(W_i - D \times K\) percent. A smartphone with this value being \(1\) or more is still usable, while a smartphone with this value less than \(1\) has powered off and become unusable.
\(K\) 天后，当充电器送达时，第 \(i\) 部手机的电量被视为 \(W_i - D \times K\) 百分比。此值大于等于 \(1\) 的手机仍可使用，而此值小于 \(1\) 的手机将关机并无法使用。
Determine the number of smartphones among the \(N\) that are still usable after \(K\) days.
确定 \(K\) 天后仍可使用的手机数量。
【输入】
\(N\) \(D\) \(K\)
\(W_1\) \(W_2\) \(\ldots\) \(W_N\)
- The first line contains the integer \(N\) representing the number of smartphones, the integer \(D\) representing the daily battery decrease, and the integer \(K\) representing the number of days until the charger arrives, separated by spaces.
- The second line contains the integers \(W_1, W_2, \ldots, W_N\) representing the current battery levels of each smartphone, separated by spaces.
【输出】
Output in one line the number of smartphones that are still usable after \(K\) days.
【输入样例】
3 10 5
100 50 30
【输出样例】
1
【解题思路】
【代码详解】
#include <bits/stdc++.h>
using namespace std;
#define int long long
int n, d, k, ans; // n: 总数量，d: 每天消耗量，k: 天数，ans: 结果计数signed main()
{cin >> n >> d >> k; // 读入n, d, kfor (int i = 1; i <= n; i++) // 遍历每个物品{int w; cin >> w; // 读入当前物品的初始重量w = w - d * k; // 计算经过k天后的剩余重量if (w >= 1) // 如果剩余重量大于等于1{ans++; // 计数加1}}cout << ans << endl; // 输出满足条件的物品数量return 0;
}
【运行结果】
3 10 5
100 50 30
1