#include <stdio.h>
#include <math.h>

int main() {
    int c, k, g;
    scanf("%d %d %d", &c, &k, &g);
    int a[21][21][21] = {0};
    int n;
    int c1, c2, k1, k2, g1, g2;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d %d %d %d %d %d", &c1, &k1, &g1, &c2, &k2, &g2);
        for (int m = c1; m <= c2; m++) {
            for (int j = k1; j <= k2; j++) {
                for (int l = g1; l <= g2; l++) {
                    a[m][j][l] = 1;
                }
            }
        }
    }
    int count = 0;
    for (int i = 0; i < c; i++) {
        for (int j = 0; j < k; j++) {
            for (int l = 0; l < g; l++) {
                if (a[i][j][l] == 0) {
                    count++;
                }
            }
        }
    }
    printf("%d\n", count);
    return 0;
}
想法是没挖的为0 挖掉的为1  然后遍历数组算剩下的个数