思路: 数学题

v, x, y = list(map(int, input().split()))print (v // x * y)

思路: 枚举

其实横竖都有dfs字符，只有3种情况

• 第一行，第一列为dfs
• 第二行，第二列为dfs
• 第三行，第三列为dfs

枚举取最小代价即可

grids = []
for i in range(3):grids.append(input())# 枚举
def cost(y, x, ch) -> int:if grids[y][x] == ch:return 0return 1r1 = cost(0, 0, 'd') + cost(0, 1, 'f') + cost(0, 2, 's') \+ cost(1, 0, 'f') + cost(2, 0, 's')r2 = cost(1, 0, 'd') + cost(1, 1, 'f') + cost(1, 2, 's') \+ cost(0, 1, 'd') + cost(2, 1, 's')r3 = cost(2, 0, 'd') + cost(2, 1, 'f') + cost(2, 2, 's') \+ cost(0, 2, 'd') + cost(1, 2, 'f')print (min(r1, r2, r3))

思路: 贪心

猜结论，感觉就是排序后

累加 abs(i - arr[i])

n = int(input())arr = list(map(int, input().split()))
arr.sort()res = 0
for i in range(n):res += abs(i + 1 - arr[i])print (res)

思路: 树形DP

自底向上的DFS即可

n = int(input())
s = input()g = [[] for _ in range(n + 1)]for i in range(n - 1):u, v = list(map(int, input().split()))g[u].append(v)g[v].append(u)from types import GeneratorTypedef bootstrap(f, stack=[]):def wrappedfunc(*args, **kwargs):if stack:return f(*args, **kwargs)else:to = f(*args, **kwargs)while True:if type(to) is GeneratorType:stack.append(to)to = next(to)else:stack.pop()if not stack:breakto = stack[-1].send(to)return toreturn wrappedfunc# python dfs会栈溢出
dp = [0] * (n + 1)@bootstrap
def dfs(u: int, fa: int):for v in g[u]:if v == fa:continueyield dfs(v, u)dp[u] += dp[v]if s[u - 1] == '1':dp[u] += 1yielddfs(1, -1)for i in range(1, n + 1):print (dp[i] - (1 if s[i - 1] == '1' else 0))

思路: 前缀和 + 异或map技巧

就是0~9这10个构建一个字符集

由于奇偶特性可以借助异或来表达

这样就变成1024种状态

时间复杂度为

$O(10\ast n)$

x = input()from collections import Countercnt = Counter()
cnt[0] = 1res = 0
s = 0
for c in x:p = ord(c) - ord('0')s ^= (1 << p)res += cnt[s]for i in range(10):res += cnt[s ^ (1 << i)]cnt[s] += 1print (res)

思路: 状压 + 3进制

很典的一道状压入门题

时间复杂度:

$O(3^{2n}\ast m)$

n, m = list(map(int, input().split()))def mapping(c) -> int: if c == 'r': return 0elif c == 'e':return 1return 2grids = []
for i in range(n):s = input()grids.append([mapping(c) for c in s])# 状压DP
# 3进制from math import pow, inf
y = int(pow(3, n))dp = [0] * ydef compute(idx, state) -> int:if not isValid(state):return infcost = 0for i in range(n):d = state % 3state = state // 3if grids[i][idx] != d:cost += 1return costdef isValid(state) -> bool:r = []for i in range(n):r.append(state % 3)state = state // 3for i in range(len(r) - 1):if r[i] == r[i+1]:return Falsereturn Truedef twoValid(s1: int, s2: int) -> bool:r1, r2 = [], []for i in range(n):r1.append(s1 % 3)s1 = s1 // 3r2.append(s2 % 3)s2 = s2 // 3for i in range(len(r1) - 1):if r1[i] == r1[i+1] or r2[i] == r2[i+1]:return Falsefor i in range(len(r1)):if r1[i] == r2[i]:return Falsereturn Truefor i in range(y):dp[i] = compute(0, i)for j in range(1, m):dp2 = [inf] * yfor k in range(y):c = compute(j, k)for k2 in range(y):if twoValid(k, k2):dp2[k] = min(dp2[k], dp[k2] + c)dp = dp2print(min(dp))