这个挑战真的很简单（并且是更困难挑战的先兆！）。
给定一个资源访问数组（用非负整数简单表示）和一个参数n，返回假定我们的缓存具有容量n并在充满时使用先进先出（FIFO）弹出方案的缓存未命中数。
例：
4, [0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 0, 1, 2, 3]
0 = not in cache (miss), insert, cache is now [0]
1 = not in cache (miss), insert, cache is now [0, 1]
2 = not in cache (miss), insert, cache is now [0, 1, 2]
3 = not in cache (miss), insert, cache is now [0, 1, 2, 3]
0 = in cache (hit), cache unchanged
1 = in cache (hit), cache unchanged
2 = in cache (hit), cache unchanged
3 = in cache (hit), cache unchanged
4 = not in cache (miss), insert and eject oldest, cache is now [1, 2, 3, 4]
0 = not in cache (miss), insert and eject oldest, cache is now [2, 3, 4, 0]
0 = in cache (hit), cache unchanged
1 = not in cache (miss), insert and eject oldest, cache is now [3, 4, 0, 1]
2 = not in cache (miss), insert and eject oldest, cache is now [4, 0, 1, 2]
3 = not in cache (miss), insert and eject oldest, cache is now [0, 1, 2, 3]
因此，在此示例中，有9个未中。也许代码示例有助于更好地解释它。在Python中：
def num_misses(n, arr):
misses = 0
cache = []
for access in arr:
if access not in cache:
misses += 1
cache.append(access)
if len(cache) > n:
cache.pop(0)
return misses其他一些测试用例（其中包含对下一个挑战的提示-注意到任何好奇的地方吗？）：
0, [] -> 0
0, [1, 2, 3, 4, 1, 2, 3, 4] -> 8
2, [0, 0, 0, 0, 0, 0, 0] -> 1
3, [3, 2, 1, 0, 3, 2, 4, 3, 2, 1, 0, 4] -> 9
4, [3, 2, 1, 0, 3, 2, 4, 3, 2, 1, 0, 4] -> 10
以字节为单位的最短代码获胜。
@JungHwanMin正确！实际上，它变得更糟的程度是无限的。
—
orlp
我们可以用一元输出数字吗？
—
dylnan '18
@dylnan不，对不起。
—
orlp
notice anything curious?一段时间，只是注意到了，增加缓存容量并不一定减少未命中次数？！