题型: 签到

贪心即可，所以值往最大值靠拢即可

arr = list(map(int, input().split()))z = max(arr)res = 0
for v in arr:res += (z - v)print (res)

思路: 暴力

暴力枚举+模拟即可

s = input()def compute(cs):res = 0i, j = 0, len(cs) - 1while i < j:if cs[i] != cs[j]:res += 1i+=1j-=1return resn = len(s)
res = 0
for i in range(n):for j in range(i, n):res += compute(s[i:j+1])print (res)

思路: 模拟

模拟即可，顺序执行相同操作

• 最后一个元素不等，即无解
• 最后一个元素相等，即有解，且按序操作最优

n = int(input())
s1 = input()
s2 = input()res = []
ss1 = list(s1)
ss2 = list(s2)for i in range(n - 1):if ss1[i] != ss2[i]:res.append([i+1, i+2])ss1[i] = '0' if ss1[i] == '1' else '1'if i + 1 < n:ss1[i+1] = '0' if ss1[i+1]=='1' else '1'if ss1[-1] != ss2[-1]:print (-1)
else:print (len(res))for (a, b) in res:print (a, b)

思路: 枚举+前后缀拆解

可以枚举除2的点，然后快速寻找那个那个点乘2收益最大

所以它的思路非常的直接

• 预处理x2的结果
• 枚举除2的点

由于不确定x2的点的位子在枚举点那一侧，所以引入前后缀拆解

这题绕的地方在于， x2和除2太接近，彼此会互相影响

所以这里需要打个补丁，即在(-1， 0， 1)的位子，特殊枚举x2的结果

这就是大致的思路

时间复杂度为$O(n)$

n = int(input())
arr = list(map(int, input().split()))def evalute(arr):res = 0n = len(arr)for i in range(n - 1):res += abs(arr[i] - arr[i + 1])return respre = [0] * n
suf = [0] * ndef gain(i):res = 0if i >= 1:res += abs(arr[i] * 2 - arr[i - 1]) - abs(arr[i] - arr[i - 1])if i < n - 1:res += abs(arr[i] * 2 - arr[i + 1]) - abs(arr[i] - arr[i + 1])return resdef divgain(i):res = 0if i >= 1:res += abs(arr[i] // 2 - arr[i - 1]) - abs(arr[i] - arr[i - 1])if i < n - 1:res += abs(arr[i] // 2 - arr[i + 1]) - abs(arr[i] - arr[i + 1])return resfrom math import inf
pre[0] = inf
for i in range(1, n):pre[i] = min(pre[i - 1], gain(i - 1))suf[n - 1] = inf
for i in range(n - 2, -1, -1):suf[i] = min(suf[i + 1], gain(i + 1))# 前后缀拆解
from math import inf
res = inf
now = evalute(arr)def recompute(arr, i):acc = 0for k in (-1, 0, 1, 2):if i + k >= 1 and i + k < n:acc += abs(arr[i + k] - arr[i + k - 1])return accfor i in range(n):res = min(res, now + divgain(i) + min(pre[i - 1] if i >= 1 else inf, suf[i + 1] if i + 1 < n else inf))        acc = recompute(arr, i)cb = arr[i]arr[i] = cb // 2for k in (-1, 0, 1):if i + k < 0 or i + k >= n:continueold = arr[i + k]arr[i + k] = old * 2res = min(res, now + recompute(arr, i) - acc)arr[i + k] = oldarr[i] = cbprint (res)

思路: 容斥+贡献

正难则反

正面求解很难，所以试试反向解法，就是统计相同的字符对

总的配对数 - 相同配对数

而这边的相同配对(i, j)，其实有一个加权，这个加权

$$w=min(i+1,n-j)$$

可以先写一个$O(n^2)$的解法

一旦写出$O(n^2)$的解法，基本上半只脚迈入成功的道路上了

因为这里能找到一个技巧点，使得时间复杂度降为$O(N)$

s = input()
n = len(s)# 考虑贡献法
r = 0
for i in range(n):d = i + 1x = n - 1 - iif x >= i:r += d * (x - i)r += d * (d - 1) // 2else:d2 = (n - i - 1)r += (d2 + 1) * d2 // 2hash = [[] for _ in range(26)]for (i, c) in enumerate(s):p = ord(c) - ord('a')hash[p].append(i)res = 0
for i in range(26):ls = hash[i]m = len(ls)pre = [0] * (m + 1)for j in range(m):pre[j + 1] = pre[j] + (n - ls[j])# 双指针优化tmp = 0k = m - 1for j in range(m):while k >= 0 and n - ls[k] < ls[j] + 1:k -= 1if k > j:tmp += (k - j) * (ls[j] + 1)tmp += pre[m] - pre[k + 1]else:tmp += pre[m] - pre[j + 1]res += tmp# 容斥
print (r - res)

思路: 引入虚节点 + 0-1BFS

• 虚节点为质数
• 节点到虚节点代价为0
• 虚节点到节点代价为1

然后就跑一下0-1BFS，即可

h, w = list(map(int, input().split()))g = []
for _ in range(h):row = list(map(int, input().split()))g.append(row)from math import inf
dp = [[inf] * w for _ in range(h)]from collections import deque
from collections import defaultdict# 建图
cnt = defaultdict(list)es = [[[] for _ in range(w)] for _ in range(h)]for i in range(h):for j in range(w):k = 2v = g[i][j]while k * k <= v:if v % k == 0:cnt[k].append((i, j))es[i][j].append(k)while v % k == 0:v = v // kk += 1if v > 1:cnt[v].append((i, j))es[i][j].append(v)deq = deque()
deq.append((0, 0, 0, 0))
dp[0][0] = 0from collections import Counter
opt = Counter()from math import gcdwhile len(deq) > 0:op, y, x, c = deq.popleft()if op == 0:if y + 1 < h:if dp[y + 1][x] > c + 1:dp[y + 1][x] = c + 1deq.append((0, y+ 1, x, c + 1))if x + 1 < w:if dp[y][x + 1] > c + 1:dp[y][x + 1] = c + 1deq.append((0, y, x + 1, c+ 1))for v in es[y][x]:if v not in opt:opt[v] = cdeq.appendleft((1, v, 0, c))else:for (ty, tx) in cnt[y]:if dp[ty][tx] > c + 1:dp[ty][tx] = c + 1deq.append((0, ty, tx, c + 1))print (dp[-1][-1])