【題解】ZeroJudge b840: 104北二4.農作物採收問題

#include <iostream>
using namespace std;
 
int n, arr[25][25], maxi, temp;
 
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= n; j++){
            cin >> arr[i][j];
        }
    }
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= n; j++){
            arr[i][j] += arr[i-1][j] + arr[i][j-1] - arr[i-1][j-1];
        }
    }
        
    maxi = 0;
    for (int a = 1; a <= n; a++){
        for (int b = 1; b <= n; b++){
            for (int c = 0; c < a; c++){
                for (int d = 0; d < b; d++){
                    temp = arr[a][b]-arr[a][d]+arr[ c][d]-arr[ c][b];
                    if (temp > maxi) maxi = temp;
                }
            }
        }
    }
    cout << maxi << "\n";
}

#測資有不預期的換行符號，須自行處理
tmp = []
while True:
    try:
        tmp.extend(list(map(int, input().split())))
    except:
        break

n = tmp[0]
tmp =  tmp[1:]
#把讀入的測資轉換成 n*n 的陣列
a = [[0 for _ in range(n+1)]]
for i in range(n):
    a.append([0]+tmp[i*n:(i+1)*n])

for i in range(1, n+1):
    for j in range(1, n+1):
        a[i][j] += a[i-1][j] + a[i][j-1] - a[i-1][j-1]

mx = 0
for i in range(1, n+1):
    for j in range(1, n+1):
        for k in range(i):
            for l in range(j):
                tmp = a[i][j] - a[i][l] - a[k][j] + a[k][l]
                mx = max(mx, tmp)
print(mx)