codeforces 1187F. Expected Square Beauty

求$E({B(x)}^2)$，考虑$B(x)$为每一位与前一位不同的期望次数

令$A(x)$表示第$x$位与第$x-1$位不同的概率，特别地，$A(1)=1$

$$E({B(x)}^2)=E({(\sum_{i=1}^n A(i))}^2)$$

把式子展开得，

$$E({B(x)}^2)=\sum_{i=1}^n \sum_{j=1}^n E(A(i)\times A(j))$$

显然如果$|i-j|>1$，$A(i)$与$A(j)$是独立的，$E(A(i)\times A(j))=E(A(i))\times E(A(j))$

$A(x)$可以在$O(1)$的时间内算出来，$|i-j|>1$的情况也可以快速计算出来。

若$|i-j|=0$，$E({A(i)}^2)=E(A(i))$。

剩下的我们就要统计$a_{i-1}\neq a_{i}且a_{i}\neq a_{i+1}$这样的概率。

可以用容斥$O(1)$计算出来。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 #define MOD 1000000007
 5 int l[200010], r[200010];
 6 int P[200010], sum[200010];
 7 inline int Power(int x, int y) {
 8     int ret = 1;
 9     while(y) {
10         if(y & 1) ret = 1ll * ret * x % MOD;
11         x = 1ll * x * x % MOD; y >>= 1;
12     }
13     return ret;
14 }
15 inline int min(int a, int b, int c) {
16     return min(a, min(b, c));
17 }
18 inline int max(int a, int b, int c) {
19     return max(a, max(b, c));
20 }
21 inline int calc(int l, int r) {
22     if(l > r) return 0;
23     int ret = sum[r] - sum[l - 1] + MOD;
24     if(ret >= MOD) ret -= MOD;
25     return ret;
26 }
27 inline void Add(int &x, int y) {
28     x += y;
29     if(x >= MOD) x -= MOD;
30 }
31 int main(){
32     int n;
33     scanf("%d", &n);
34     for(int i = 1; i <= n; ++ i) {
35         scanf("%d", &l[i]);
36         -- l[i];
37     }
38     for(int i = 1; i <= n; ++ i) {
39         scanf("%d", &r[i]);
40     }
41     P[1] = 1;
42     for(int i = 2; i <= n; ++ i) {
43         int B = 1ll * (r[i - 1] - l[i - 1]) * (r[i] - l[i]) % MOD;
44         int A = min(r[i - 1], r[i]) - max(l[i - 1], l[i]);
45         if(A < 0) A = 0;
46         P[i] = 1ll * (B - A + MOD) * Power(B, MOD - 2) % MOD;
47     }
48     for(int i = 1; i <= n; ++ i) {
49         sum[i] = sum[i - 1] + P[i];
50         if(sum[i] >= MOD) sum[i] -= MOD;
51     }
52     int Ans = 0;
53     for(int i = 1; i <= n; ++ i) {
54         Add(Ans, P[i]);
55         Add(Ans, 1ll * P[i] * calc(1, i - 2) % MOD);
56         Add(Ans, 1ll * P[i] * calc(i + 2, n) % MOD);
57     }
58     for(int i = 2; i <= n; ++ i) {
59         if(i == 2) {
60             Add(Ans, 2 * P[i] % MOD);
61             continue;
62         }
63         int E = 1;
64         E -= (1 - P[i] + MOD);
65         if(E < 0) E += MOD;
66         E -= (1 - P[i - 1] + MOD);
67         if(E < 0) E += MOD;
68         int B = 1ll * (r[i - 2] - l[i - 2]) * (r[i - 1] - l[i - 1]) % MOD * (r[i] - l[i]) % MOD;
69         int A = min(r[i - 2], r[i - 1], r[i]) - max(l[i - 2], l[i - 1], l[i]);
70         if(A < 0) A = 0;
71         E += 1ll * A * Power(B, MOD - 2) % MOD;
72         if(E >= MOD) E -= MOD;
73         Add(Ans, E * 2 % MOD);
74     }
75     printf("%d\n", Ans);
76 }

原文地址：https://www.cnblogs.com/iamqzh233/p/11137506.html