陷入了“会做，但不完全会做”的状况
C
水题，排序找中间数两边的差值
D
组合数学
求把n个相同的球分到m个相同的盒子，1每个盒子至少一个球2每个盒子球不限的组合数
空挡插隔板法，高中数学

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharm
import bisect
import copy
import sys
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)mod = 10 ** 9 + 7def pw(a, x):if x == 0:return 1temp = pw(a, x >> 1)if x & 1:return temp * temp * a % modelse:return temp * temp % moddef inv(x):return pw(x, mod - 2)def main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn, k = map(int, fp.readline().split())R, B = n - k, kdef cmb(x, y):if x < y:return 0return fac[x] * inv(fac[y]) * inv(fac[x - y]) % modfac = [1] * 5005for i in range(2, 5005):fac[i] = fac[i - 1] * i % modfor i in range(1, B + 1):ans_r = cmb(R + 1, i)ans = cmb(B - 1, i - 1) * ans_r % modprint(ans)if __name__ == "__main__":main()

E
一开始用dfs去求（3步能访问到的点对），明显TLE
分层图求最短路
原图分为3层，每个点拆成$V_0,V_1,V_2$，
假如$(u,v)$间有一条有向边
把边拆成$(u_0,v_1),(u_1,v_2),(u_2,v_3)$
走三的倍数步能回到0层，这样只要能走到$T_0$点就有解

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharm
import bisect
import copy
import sys
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)def main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn, m = map(int, fp.readline().split())g = [[] for _ in range(n * 3)]for i in range(m):u, v = map(int, fp.readline().split())u, v = u - 1, v - 1u0, u1, u2 = u, u + n, u + 2 * nv0, v1, v2 = v, v + n, v + 2 * ng[u0].append(v1)g[u1].append(v2)g[u2].append(v0)S, T = map(int, fp.readline().split())S, T = S - 1, T - 1vis = [-1] * (3 * n)qu = deque()qu.append(S)vis[S] = 0while qu:cur = qu.popleft()if cur == T:breakfor v in g[cur]:if vis[v] == -1:vis[v] = vis[cur] + 1qu.append(v)if vis[T] == -1:print(-1)else:print(vis[T] // 3)if __name__ == "__main__":main()

F
分块
重要的是想清楚每一块代表的数字