題目要求

P5638題目鏈接

分析

交了兩版本優化的Java代碼，全T了最后5個點。

第一版是用了前綴和來寫的：

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int num = scanner.nextInt(), radius = scanner.nextInt();long[] sum_array = new long[num];for (int i = 0; i < num-1; i++) {sum_array[i+1] = sum_array[i] + scanner.nextLong();}scanner.close();long count = sum_array[radius];for(int i = 1; i < num-radius; i++) {count = Math.max(count, (sum_array[i+radius]-sum_array[i]));}System.out.println(sum_array[num-1]-count);} }

第二個版本用了DP來寫的（參考了大佬的講解）：

import java.util.Scanner;public class Main {public static void main(String[] args) { // Scanner scanner = new Scanner(System.in); // int num = scanner.nextInt(), radius = scanner.nextInt(); // long[] sum_array = new long[num]; // for (int i = 0; i < num-1; i++) { // sum_array[i+1] = sum_array[i] + scanner.nextLong(); // } // scanner.close(); // long count = sum_array[radius]; // for(int i = 1; i < num-radius; i++) { // count = Math.max(count, (sum_array[i+radius]-sum_array[i])); // } // System.out.println(sum_array[num-1]-count);Scanner scanner = new Scanner(System.in);int num = scanner.nextInt(), radius = scanner.nextInt();long[][] dp = new long[num+1][2];long[] array = new long[num];for (int i = 1; i < num; i++) {array[i] = scanner.nextLong();}scanner.close();for(int i = 2; i <= num;i++) {dp[i][0] = dp[i-1][0] + array[i-1];if(i > radius) {dp[i][1] = dp[i-radius][0];}dp[i][1] = Math.min(dp[i-1][1]+array[i-1], dp[i][1]);//狀態轉移方程}System.out.println(dp[num][1]);} }

AC代碼（C++語言描述）

#include<bits/stdc++.h> using namespace std; long long sum_array[1000001]; long long num, radius;int main() {cin >> num >> radius;for(int i = 0; i < num-1; i++) {long long x;cin>>x;sum_array[i+1] = sum_array[i] + x;}long long count = sum_array[radius];for(int i = 1; i < num - radius; i++) {count=max(count, sum_array[i + radius] - sum_array[i]);}cout << sum_array[num-1]-count << endl;return 0; }

總結

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