ZeroJudge 解題筆記：h495. 農家樂(Agricola)

作者：王一哲
日期：2025年11月13日


ZeroJudge 題目連結：h495. 農家樂(Agricola)

解題想法

由於各項目的分數與數量沒有特定的數學關係，我只想到對每個項目分別建一個串列，記錄各數量對應的分數。

Python 程式碼

使用時間約為 0.8 s，記憶體約為 3.6 MB，通過測試。

field = (-1, -1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)
enclosure = (-1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)
wheat = (-1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)
vegetable = (-1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)
sheep = (-1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)
pig = (-1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)
catle = (-1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)
t = int(input())
for _ in range(t):
    arr = tuple(map(int, input().split()));
    score = field[arr[0]] + enclosure[arr[1]] + wheat[arr[2]] + vegetable[arr[3]] + \
            sheep[arr[4]] + pig[arr[5]] + catle[arr[6]] - arr[7] + arr[8] + arr[9] + \
            2*arr[10] + 3*arr[11] + arr[12]
    print(score)

C++ 程式碼

使用時間約為 93 ms，記憶體約為 48 kB，通過測試。

#include <cstdio>

int main() {
    int field[21] = {-1, -1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4};
    int enclosure[21] = {-1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4};
    int wheat[21] = {-1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4};
    int vegetable[21] = {-1, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4};
    int sheep[21] = {-1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4};
    int pig[21] = {-1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4};
    int catle[21] = {-1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4};
    int t; scanf("%d", &t);
    while(t--) {
        int arr[13];
        for(int i=0; i<13; i++) scanf("%d", &arr[i]);
        int score = field[arr[0]] + enclosure[arr[1]] + wheat[arr[2]] + vegetable[arr[3]] +
                    sheep[arr[4]] + pig[arr[5]] + catle[arr[6]] - arr[7] + arr[8] + arr[9] +
                    2*arr[10] + 3*arr[11] + arr[12];
        printf("%d\n", score);
    }
    return 0;
}