Educational Codeforces Round 68 (Rated for Div. 2) F. Crossword Expert

题意：给定时间T,事件花费的时间数组a, 每件事有二分之一的几率多花费一秒钟, 求在给定时间内能晚场的事件个数的期望

题面

Strategy: 先看大佬的博客,这题有两个难点, 首先是想到该期望应该等于$$\frac{1}{2^x}\sum _{i=0}^{i\le maxdo}{C}_x^i(前x件事中有i件要多花一秒钟\times 多花一秒钟的概率)$$.然后找到一个分界点X, 使得所有在X之前的事件即使每件事都多花一秒钟也全部做得完(即对答案贡献为1的事件),然后在考虑不一定能做得完的事件, 最后想到该怎样优化组合数前缀和

要想O(1)转移组合数前缀和, 要知道一下前置知识点

• $C_n^k+C_n^{k+1}=C_{n+1}^{k+1}$
  于是聪明的你说不定会想到
• $$\sum _{i=0}^{k+1}{C}_{n+1}^i=C_n^{k+1}+2\times \sum _{i=0}^k{C}_n^i$$

于是就有以下产物

#include 
#include 
#define oo INT_MAX
#define ll long long
#define db double
#define all(a) a.begin(), a.end()
#define met(a, b) memset(a, b, sizeof(a))
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define _rev(i, a, b) for (int i = (a); i >= (b); --i)
#define _for(i, a, b) for (int i = (a); i < (b); ++i)
#define lowbit(x) x &(-x)
#define pi acos(-1.0)
using namespace std;
using namespace __gnu_pbds;
const int maxn = 2e5 + 10, mod = 1e9 + 7;
ll T, n, sumt[maxn], inv[maxn], fac[maxn], t[maxn], invf[maxn];
void cala_inv() //逆元相关
{
	invf[0] = invf[1] = fac[1] = fac[0] = 1;
	inv[1] = 1;
	_for(i, 2, maxn)
		fac[i] = fac[i - 1] * i % mod,
		inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod,
		invf[i] = invf[i - 1] * inv[i] % mod;
}

ll C(ll m, ll n)
{
	if (n < m)
		return 0;
	return (fac[n] * invf[n - m] % mod) * invf[m] % mod;
}
ll curC = 0, dx1;
ll calc_SigmaC(int cur)
{
	dx1 = min(1ll * cur, T - sumt[cur]);
	_rep(i, 0, dx1)
	{
		curC = (curC + C(i, cur)) % mod;
	}
	
	return curC;
}
ll calc_nxt(int cur)
{
	ll dx2 = min(1ll * cur, T - sumt[cur]);
	ll tmpC = ((curC * 2 % mod) + C(dx1 + 1, cur - 1)) % mod;

	while (dx1 >= dx2)
	{
		tmpC = (tmpC - C(dx1-- + 1, cur) + mod) % mod;
	}
	dx1++;
	
	// curC = tmpC, dx1 = dx2;
	return curC = tmpC;
}

int main()
{

	ios::sync_with_stdio(0);
	cin >> n >> T;
	cala_inv();
	_rep(i, 1, n)
	{
		cin >> t[i];
		sumt[i] = sumt[i - 1] + t[i];
	}
	ll ans = 0;
	int x = 1;
	ll inv2x = inv[2];
	while(x <= n && sumt[x] + x <= T){
		x ++;
		ans ++;
		inv2x = inv2x*inv[2] %mod;
	}
	if (x <= n)
	{
		calc_SigmaC(x);
		for (; x <= n; x++)
		{
			ans = (ans + curC * inv2x) % mod;
			if (T - sumt[x + 1] < 0)
				break;
			inv2x = (inv2x * inv[2]) % mod;
			curC = calc_nxt(x + 1);
		}
	}
	cout << ans << endl;
}