类型: 签到

w1,w2 = input().split()
print (w2)
print (w1)

思路: 模拟

可以从拆包的角度去构建模拟

注意拆一包，可以烧好几次的情况

n, x, k = list(map(int, input().split()))res = []
pack = 0
left = 0
for i in range(n):left += xpack += 1while left >= k:res.append(pack)left -= kpack = 0print (len(res))
print (*res, sep=' ')

思路: 枚举

其实只能删除第1/第2，只能保证删除数组的前面一个区间，最多保留一个

题外话

如果题目改成第1/第2/…/第k，那最大价值数组是多少？

其实也是类似的思路

枚举最后一个被删除的位子

这样价值为 [前缀最多一位] + [后缀区间价值和]

s = input()n = len(s)n0, n1 = 0, 0
for c in s:if c == '0':n0 += 1else:n1 += 1# 然后枚举
res = max(0, n1 - n0)t1 = 0
for i in range(n):if s[i] == '0':n0 -= 1else:n1 -= 1t1 += 1res = max(res, n1 - n0 + (1 if t1 > 0 else 0))print (res)

思路: 贪心

尽量从最小值开始删除，带动连续数组

貌似这边for嵌套while，看似时间复杂度很高，但实际上，每次都会有效删除至少1，而n最多1e5，所以均摊下来为O(1)

时间复杂度为$O(n)$

# 贪心
n = int(input())
arr = list(map(int, input().split()))from collections import Countercnt = Counter(arr)brr = [[k, v] for (k, v) in cnt.items()]
brr.sort(key=lambda x: x[0])ans = 0
m = len(brr)
for i in range(m):if brr[i][1] > 0:j = i + 1cost = brr[i][1]brr[i][1] = 0ans += costwhile j < m and cost > 0 and brr[j - 1][0] + 1 == brr[j][0]:cost = min(cost, brr[j][1])            brr[j][1] -= costj += 1print (ans)

思路: dijkstra最短路

题型: 板子题

t = int(input())import heapq
from math import infdef solve(grids, health):h, w = len(grids), len(grids[0])sy, sx = -1, -1ty, tx = -1, -1for i in range(h):for j in range(w):if grids[i][j] == 'S':sy, sx = i, jelif grids[i][j] == 'T':ty, tx = i, jif sy == -1 or ty == -1:return "No"arr = []vis = [[inf] * w for _ in range(h)]heapq.heappush(arr, (0, sy, sx))vis[sy][sx] = 0while len(arr) > 0:cur = heapq.heappop(arr)if cur[0] > vis[cur[1]][cur[2]]:continueif cur[0] >= health:continueif cur[1] == ty and cur[2] == tx:return "Yes"for (dy, dx) in [(1, 0), (-1, 0), (0, -1), (0, 1)]:gy = cur[1] + dygx = cur[2] + dxif gy >= h or gy < 0 or gx >= w or gx < 0:continuetcost = 0if grids[gy][gx] >= '0' and grids[gy][gx] <= '9':tcost = ord(grids[gy][gx]) - ord('0')if vis[gy][gx] > cur[0] + tcost and cur[0] + tcost < health:vis[gy][gx] = cur[0] + tcostheapq.heappush(arr, (vis[gy][gx], gy, gx))return "No"for _ in range(t):n, m, h = list(map(int, input().split()))grids = []for _ in range(n):grids.append(input())print (solve(grids, h))

后期再补上