CF1187F Expected Square Beauty

Expected Square Beauty

有一个长度为 n 的数列，第 i 个数的取值范围为 ([l_i,r_i]) ，定义一个数列的价值为这个数列极长连续相同段的个数，求一个数列价值的平方期望，对 (10^9+7) 取模 。

n≤200000 。

题解

https://codeforces.com/blog/entry/68111

As usual with tasks on an expected value, let's denote (I_i(x)) as indicator function: (I_i(x) = 1) if (x_i eq x_{i - 1}) and (0) otherwise; (I_1(x) = 1). Then we can note that (B(x) = sumlimits_{i = 1}^{n}{I_i(x)}). Now we can make some transformations:

[E(B^2(x)) = E((sumlimits_{i = 1}^{n}{I_i(x)})^2) = E(sumlimits_{i = 1}^{n}{ sumlimits_{j = 1}^{n}{I_i(x) I_j(x)} }) = sumlimits_{i = 1}^{n}{ sumlimits_{j = 1}^{n}{E(I_i(x) I_j(x))} } ]

Now we'd like to make some casework:

• if |i - j| > 1 (i and j aren't consecutive) then (I_i(x)) and (I_j(x)) are independent, that's why (E(I_i(x) I_j(x)) = E(I_i(x)) E(I_j(x)));

• if i = j then (E(I_i(x) I_i(x)) = E(I_i(x)));

• |i - j| = 1 need further investigation.

For the simplicity let's transform segment ([l_i, r_i]) to ([l_i, r_i)) by increasing (r_i = r_i + 1).

Let's denote (q_i) as the probability that (x_{i-1} = x_i):

[q_i = max(0, frac{min(r_{i - 1}, r_i) - max(l_{i - 1}, l_i)}{(r_{i - 1} - l_{i - 1})(r_i - l_i)}) ]

and (q_1 = 0). Let's denote (p_i = 1 - q_i). In result, (E(I_i(x)) = p_i).

The final observation is the following: (E(I_i(x) I_{i + 1}(x))) is equal to the probability that (x_{i - 1} eq x_i) and (x_i eq x_{i + 1}) and can be calculated by inclusion-exclusion principle:

[E(I_i(x) I_{i + 1}(x)) = 1 - q_{i} - q_{i + 1} + P(x_{i-1} = x_i & x_i = x_{i + 1}) ]

, where

[P(x_{i-1} = x_i & x_i = x_{i + 1}) = max(0, frac{min(r_{i-1}, r_i, r_{i+1}) - max(l_{i-1}, l_i, l_{i+1})}{(r_{i-1} - l_{i-1})(r_i - l_i)(r_{i+1} - l_{i+1})}). ]

In result,

[E(B^2(x)) = sumlimits_{i = 1}^{n}{( p_i + p_isumlimits_{|j - i| > 1}{p_j} + E(I_{i-1}(x) I_i(x)) + E(I_i(x) I_{i+1}(x)) )} ]

and can be calculated in (O(n log{MOD})) time.

CO int N=2e5+10;
int L[N],R[N];
int Q[N],E[N];

int calc(int i,int j,int k){
	int prob=0;
	if(i>=1){
		int l=max(L[i],max(L[j],L[k])),r=min(R[i],min(R[j],R[k]));
		if(l<r) prob=mul(r-l,fpow(mul(R[i]-L[i],
					 mul(R[j]-L[j],R[k]-L[k])),mod-2));
	}
	return add(1,add(mod-Q[j],add(mod-Q[k],prob)));
}

int main(){
	int n=read<int>();
	for(int i=1;i<=n;++i) read(L[i]);
	for(int i=1;i<=n;++i) read(R[i]),++R[i];
	int sum=0;
	for(int i=1;i<=n;++i){
		int l=max(L[i],L[i-1]),r=min(R[i],R[i-1]);
		if(l<r) Q[i]=mul(r-l,fpow(mul(R[i-1]-L[i-1],R[i]-L[i]),mod-2));
		E[i]=add(1,mod-Q[i]);
		sum=add(sum,E[i]);
	}
	int ans=0;
	for(int i=1;i<=n;++i){
		int res=sum;
		for(int j=max(i-1,1);j<=min(i+1,n);++j) res=add(res,mod-E[j]);
		ans=add(ans,mul(E[i],res));
		ans=add(ans,E[i]);
		if(i>1) ans=add(ans,calc(i-2,i-1,i));
		if(i<n) ans=add(ans,calc(i-1,i,i+1));
	}
	printf("%d
",ans);
	return 0;
}