牛客题解 | 罪犯转移

题目

题目链接

解题思路

这是一个典型的滑动窗口问题：

1. 需要找出长度为 \(c\) 的连续子数组
2. 子数组的和不能超过 \(t\)
3. 统计满足条件的子数组个数

关键点

1. 使用滑动窗口维护连续 \(c\) 个元素的和
2. 注意处理多组测试数据
3. 数据范围较大，注意使用long long类型

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代码

#include <iostream>
#include <vector>
using namespace std;class Solution {
public:long long countValidTransfers(int n, long long t, int c, vector<int>& crimes) {long long count = 0;long long windowSum = 0;// 计算第一个窗口的和for (int i = 0; i < c; i++) {windowSum += crimes[i];}// 如果第一个窗口满足条件，计数加1if (windowSum <= t) {count++;}// 滑动窗口for (int i = c; i < n; i++) {// 更新窗口和：减去窗口最左边的值，加上新的值windowSum = windowSum - crimes[i-c] + crimes[i];// 检查当前窗口是否满足条件if (windowSum <= t) {count++;}}return count;}
};int main() {int n, c;long long t;// 处理多组测试数据while (cin >> n >> t >> c) {vector<int> crimes(n);for (int i = 0; i < n; i++) {cin >> crimes[i];}Solution solution;cout << solution.countValidTransfers(n, t, c, crimes) << endl;}return 0;
}

import java.util.*;public class Main {static class Solution {public long countValidTransfers(int n, long t, int c, int[] crimes) {long count = 0;long windowSum = 0;// 计算第一个窗口的和for (int i = 0; i < c; i++) {windowSum += crimes[i];}// 如果第一个窗口满足条件，计数加1if (windowSum <= t) {count++;}// 滑动窗口for (int i = c; i < n; i++) {// 更新窗口和windowSum = windowSum - crimes[i-c] + crimes[i];// 检查当前窗口是否满足条件if (windowSum <= t) {count++;}}return count;}}public static void main(String[] args) {Scanner sc = new Scanner(System.in);// 处理多组测试数据while (sc.hasNext()) {int n = sc.nextInt();long t = sc.nextLong();int c = sc.nextInt();int[] crimes = new int[n];for (int i = 0; i < n; i++) {crimes[i] = sc.nextInt();}Solution solution = new Solution();System.out.println(solution.countValidTransfers(n, t, c, crimes));}sc.close();}
}

class Solution:def count_valid_transfers(self, n: int, t: int, c: int, crimes: list) -> int:count = 0window_sum = 0# 计算第一个窗口的和for i in range(c):window_sum += crimes[i]# 如果第一个窗口满足条件，计数加1if window_sum <= t:count += 1# 滑动窗口for i in range(c, n):# 更新窗口和window_sum = window_sum - crimes[i-c] + crimes[i]# 检查当前窗口是否满足条件if window_sum <= t:count += 1return countdef main():# 处理多组测试数据while True:try:n, t, c = map(int, input().split())crimes = list(map(int, input().split()))solution = Solution()print(solution.count_valid_transfers(n, t, c, crimes))except EOFError:breakif __name__ == "__main__":main()

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算法及复杂度

• 算法：滑动窗口
• 时间复杂度：\(\mathcal{O}(n)\)，只需要遍历一次数组
• 空间复杂度：\(\mathcal{O}(1)\)，只使用常数额外空间