优化前后的两个数

运行下面这段程序，填写它打印出来的结果。

BLK = 4

def loads(idxs):
    cur = -1
    n = 0
    for i in idxs:
        b = i // BLK
        if b != cur:
            n = n + 1
            cur = b
    return n

N = 8
before = [i * N + j for j in range(N) for i in range(N)]
after = [i * N + j for i in range(N) for j in range(N)]
print(str(loads(before)) + "/" + str(loads(after)) + "/"
      + str(loads(before) // loads(after)))