CF1187F Expected Square Beauty
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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;
}
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