题目
题目链接
解题思路
-
基本思路：
- 先对题目难度进行排序。
- 从最小难度开始，尝试组合每场考试的三道题。
- 每找到一组题目后，标记这些题目为已使用。
- 统计需要补充的题目数量。
-
实现方法：
- 使用贪心策略，优先使用难度相近的题目组合。
- 使用 \(\text{used}\) 数组标记已使用的题目。
- 对于每组题目，确保相邻题目难度差不超过 \(10\)。
- 计算每组不足三道题时需要补充的题目数量。
代码
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;int minExtraQuestions(int n, vector<int>& difficulties) {sort(difficulties.begin(), difficulties.end());vector<bool> used(n, false);int extra = 0;// 从最小的数开始尝试组合for (int i = 0; i < n; i++) {if (used[i]) continue;// 找第一道题int first = difficulties[i];used[i] = true;// 找第二道题int second = -1, second_idx = -1;for (int j = i + 1; j < n; j++) {if (!used[j] && difficulties[j] - first <= 10) {second = difficulties[j];second_idx = j;break;}}// 找第三道题int third = -1, third_idx = -1;if (second != -1) {for (int j = second_idx + 1; j < n; j++) {if (!used[j] && difficulties[j] - second <= 10) {third = difficulties[j];third_idx = j;break;}}}// 计算需要补充的题目数if (second == -1) { // 只有第一道题extra += 2;} else if (third == -1) { // 只有两道题used[second_idx] = true;extra += 1;} else { // 找到三道题used[second_idx] = true;used[third_idx] = true;}}return extra;
}int main() {int n;cin >> n;vector<int> difficulties(n);for (int i = 0; i < n; i++) {cin >> difficulties[i];}cout << minExtraQuestions(n, difficulties) << endl;return 0;
}
import java.util.*;public class Main {public static int minExtraQuestions(int n, int[] difficulties) {Arrays.sort(difficulties);boolean[] used = new boolean[n];int extra = 0;// 从最小的数开始尝试组合for (int i = 0; i < n; i++) {if (used[i]) continue;// 找第一道题int first = difficulties[i];used[i] = true;// 找第二道题int second = -1, second_idx = -1;for (int j = i + 1; j < n; j++) {if (!used[j] && difficulties[j] - first <= 10) {second = difficulties[j];second_idx = j;break;}}// 找第三道题int third = -1, third_idx = -1;if (second != -1) {for (int j = second_idx + 1; j < n; j++) {if (!used[j] && difficulties[j] - second <= 10) {third = difficulties[j];third_idx = j;break;}}}// 计算需要补充的题目数if (second == -1) { // 只有第一道题extra += 2;} else if (third == -1) { // 只有两道题used[second_idx] = true;extra += 1;} else { // 找到三道题used[second_idx] = true;used[third_idx] = true;}}return extra;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int[] difficulties = new int[n];for (int i = 0; i < n; i++) {difficulties[i] = sc.nextInt();}System.out.println(minExtraQuestions(n, difficulties));}
}
def min_extra_questions(n, difficulties):difficulties.sort()used = [False] * nextra = 0# 从最小的数开始尝试组合for i in range(n):if used[i]:continue# 找第一道题first = difficulties[i]used[i] = True# 找第二道题second = -1second_idx = -1for j in range(i + 1, n):if not used[j] and difficulties[j] - first <= 10:second = difficulties[j]second_idx = jbreak# 找第三道题third = -1third_idx = -1if second != -1:for j in range(second_idx + 1, n):if not used[j] and difficulties[j] - second <= 10:third = difficulties[j]third_idx = jbreak# 根据找到的题目数量计算需要补充的题目if second == -1: # 只有第一道题extra += 2elif third == -1: # 只有两道题used[second_idx] = Trueextra += 1else: # 找到三道题used[second_idx] = Trueused[third_idx] = Truereturn extraif __name__ == "__main__":n = int(input())difficulties = list(map(int, input().split()))print(min_extra_questions(n, difficulties))
算法及复杂度
- 算法：贪心算法
- 时间复杂度：\(\mathcal{O}(n^2)\)，其中 \(n\) 为题目数量
- 空间复杂度：\(\mathcal{O}(n)\)，用于存储已使用题目的标记
