C - Rally
从0到100模拟一遍位置最小的点
最小值实际上可以计算导数，将复杂度降为$O(1)$

D - Bouquet
组合数$2^n-1-C(n,a)-C(n,b)$
由于a.b都不大，因此可以做。注意用递推计算逆元.

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(1000)mod = 10 ** 9 + 7def pw(a, x):if x == 1:return atemp = pw(a, x >> 1)if x & 1:return temp * temp * a % modelse:return temp * temp % moddef main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn, a, b = map(int, fp.readline().split())t0 = pw(2, n) - 1m = max(a, b) + 1fac = [1] * mfor i in range(1, m):fac[i] = fac[i - 1] * i % modinv = [1] * minv[m - 1] = pw(fac[m - 1], mod - 2)for i in range(m - 2, -1, -1):inv[i] = inv[i + 1] * (i + 1) % moddef cmb(x, y):ret = inv[y]for i in range(x, x - y, -1):ret = ret * i % modreturn retans = (t0 - cmb(n, a) - cmb(n, b)) % modprint(ans)if __name__ == "__main__":main()

E - Roaming
组合题
由于原始数组中每个数都为1，那么操作等价于将1变为0，然后看剩下的一些位置中如何放球。
枚举0的个数，设有i个位置为0，那么剩下的n-i位置中要安放n个球，且每个位置至少放1个球，
a个球放在b位置中每个位置至少一个球的方案数等价于在a个球的空隙中插b-1个隔板，因此为$C(a-1,b-1)$

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(1000)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())mod = 10 ** 9 + 7ans = 0fac = [1] * (n + 1)for i in range(1, n + 1):fac[i] = fac[i - 1] * i % moddef pw(a, x):if x == 1:return at = pw(a, x >> 1)if x & 1:return t * t * a % modelse:return t * t % modinv = [1] * (n + 1)inv[n] = pw(fac[n], mod - 2)for i in range(n - 1, -1, -1):inv[i] = inv[i + 1] * (i + 1) % moddef c(x, y):return fac[x] * inv[y] * inv[x - y] % modans = 0for i in range(k + 1):if n - i - 1 >= 0:ans += c(n - 1, n - i - 1) * c(n, i) % modans %= modelse:breakprint(ans)if __name__ == "__main__":main()