BZOJ-3450 Easy probability dp "mathematical expectation"

Description

One day WJMZBMR is playing osu~~~ but he is too weak to push, some places entirely by luck: (
Let's simplify the rules of the game.
Have n click to do, success is O, failure is x, score is calculated according to comb, continuous a comb there is a*a points, comb is a great continuous o.
Like Ooxxxxooooxxx, the score is 2*2+4*4=4+16=20.
Sevenkplus Idle panic To see he played a plate, some places have nothing to do with luck either O or x, some places o or x each have 50% probability, with the number to express.
For example, OO?XX is a possible input.
So what is the expected score of WJMZBMR this OSU?
Like Oo?xx's words, "o" is the word Oooxx = 9, x is ooxxx = 4
The natural expectation is (4+9)/2 = 6.5.

Input

The first line is an integer n, which indicates the number of clicks
Next string, each character is an ox? In one of

Output

A single line of floating-point numbers indicates the answer
Rounded to 4 digits after the decimal point
If the fear of precision kneeling suggested with a long double or extended

Sample Input4
????
Sample Output4.1250

n<=300000
Solvingtaking into account each contribution to the expectation, it is assumed that in the processing of the position I, I have a continuous o, which can be considered a special case. 　　 Suppose this bit is o then? i = (l + 1) ^ 2-l ^ 2 = 2 * l + 1, while taking expected E (? i) = 2 * E (L) + 1;Suppose this bit is x then? i = 0;
Suppose this one is? Then E (? i) = P1 * X1 + p0 * X0 (where P1 is the probability of taking one of the bits, P0 is the probability of taking 0, X1 for this bit to take 1 of the score changes, X0 the same), then X0 = 0, so e (? i) = P1 * X1,e (? i) = 0.5 * (2 * E (L) + 1) = e (L) + 0.5. then the question becomes how to solve the desired length. Suppose this bit is O then E (l) + 1;Suppose this bit is x so E (L) = 0;Suppose this bit is? Then E (l) = 0.5 * 0 + 0.5 * (E (L) + 1);　　O (n) can be processed.

1#include <bits/stdc++.h>2 #defineRep (I, A, b) for (int i = A; I <= b; i++)3 #defineDrep (I, A, b) for (int i = A; I >= b; i--)4 #defineREP (I, A, b) for (int i = A; I < b; i++)5 #definePB Push_back6 #defineMP Make_pair7 #defineCLR (x) memset (x, 0, sizeof (x))8 #defineXX First9 #defineyy secondTen using namespacestd; OnetypedefLong Longi64; Atypedef pair<int,int>PII; - Const intINF = ~0U>>1; - Consti64 INF = ~0ull >>1; the //*************************************** -   - intMain () { -     intN; +scanf"%d\n", &n); -     DoubleL0), ans (0); +     Charch; A      while(n--) { atCH =GetChar (); -         if(ch = ='o') ans + =2* L +1, l++; -         Else if(ch = ='x') L =0; -         ElseAns + = L +0.5, L = (l +1) /2; -     } -printf"%.4LF", ans); in     return 0; -}

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BZOJ-3450 Easy probability dp "mathematical expectation"