教学计划-找规律II
2024-09-29
240929julyfunnoteswilliam工作Circular Barnhttps://usaco.org/index.php?page=viewproblem2&cpid=1255首先考虑一个房间内如果一直拿，谁会赢。从 1 开始推导，发现当数量为 4k 时，后手赢，其他先手赢。需要拿几次可以取完？如果后手赢，那么先手要尽可能让对方赢所需拿的次数多一点。如果先手赢，每个房间遵循尽快赢的策略。有，若偶数，则次数为 i // 2，否则若为 4k + 1，则取最大质数满足取完后为 4k，而 4k + 3 类似: if i % 2 == 0:
towin[i] = i / 2
elif i % 4 == 1:
if once[i]:
m1 = i
towin[i] = 1
else:
towin[i] = (i - m1) / 2 + 1
elif i % 4 == 3:
if once[i]:
m3 = i
towin[i] = 1
else:
towin[i] = (i - m3) / 2 + 1首先用质数筛筛出质数。对于取完次数为 k 的房间，需要第 k / 2 + 1 轮到这个房间才能有人输。N = int(5e6)
once = [True] * (N + 1)
prime = []
for i in range(2, N + 1):
if once[i]:
prime.append(i)
for j in range(len(prime)):
v = prime[j]
if v * i > N:
break
once[v * i] = False
if i % v == 0:
break
m3 = 3 # largest i: 4i + 3 prime
m1 = 1
towin = [0, 1, 1, 1] + [0] * int(N + 10)
for i in range(4, int(N + 1)):
if i % 2 == 0:
towin[i] = i // 2
elif i % 4 == 1:
if once[i]:
m1 = i
towin[i] = 1
else:
towin[i] = (i - m1) // 2 + 1
elif i % 4 == 3:
if once[i]:
m3 = i
towin[i] = 1
else:
towin[i] = (i - m3) // 2 + 1
t = int(input())
for _ in range(t):
n = int(input())
a = [0] + list(map(int, input().split()))
firstmin = 1e9
win = 0
for i in range(1, n + 1):
rd = towin[a[i]] // 2 + 1
if rd < firstmin:
firstmin = rd
win = towin[a[i]] % 2
if win == 1:
print("Farmer John")
else:
print("Farmer Nhoj")Sleepy Cow Herdinghttps://usaco.org/index.php?page=viewproblem2&cpid=918最少的步数：假设最后所有奶牛集中到 p ~ q，显然有 a[0] <= p < q <= a[n - 1]，长度为 n，那么发现任意指定一个 p ~ q，每一步都可以把一头在外面的奶牛移进来。只需要双指针法找到长度为 n 的包含奶牛个数最多的区间就行。最多的步数：定义总距离为每对相邻奶牛之间距离 - 1 的和。如果边缘有两头奶牛相邻，发现每步一定可以只让距离减 1，而第一步一定会把两端移到中间某个位置，故最优方案就是两端先造成一个相邻，之后不断减 1n = int(input())
a = [int(input()) for _ in range(n)]
a.sort()
t = 0
minans = 1e9
for s in range(n):
while t + 1 <= n - 1 and a[t + 1] - a[s] + 1 <= n:
t += 1
minans = min(minans, n - (t - s + 1))
print(minans)
totdis = a[n - 1] - a[0] - (n - 1)
print(totdis - min(a[2] - a[1], a[n - 1] - a[n - 2]) + 1)作业题:COW Operationshttps://usaco.org/index.php?page=viewproblem2&cpid=123