Bzoj 3450:tyvj1952 Easy expectation/probability, dynamic planning

3450:tyvj1952 Easy Time limit:10 Sec Memory limit:128 MB
submit:431 solved:325
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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
OSU, it's fun.
WJMZBMR technology is OK (fog), X is basically very few
HINT Source

We all love the Gyz Cup.

Exercises

Probability/expectation + dynamic planning

Set L to the length of the desired O

When the next number is X, the l=0

When the next number is O, l++, ans=ans+ (l+1) * (l+1)-l*l

=ans+l^2+2*l+1-l^2

=ans+2*l+1

When the next number is, L may be 0 or l+1, so L's expected value is (0+l+1)/2

In the same vein, ans may add 0 or 2*l+1, so the desired value of ans is added to the number of (0+2*l+1)/2

Then you can move the rules directly.

1#include <bits/stdc++.h>2 using namespacestd;3 intMain ()4 {5     intn,i;6     Doubleans=0.0, l=0.0;7     CharA;8scanf"%d\n",&n);9      for(i=1; i<=n;i++)Ten     { Onescanf"%c",&a); A         if(a=='o') {ans+= (l*2.0+1.0); l++;} -         Else if(a=='x') l=0.0; -         Else{ans=ans+ (l*2.0+1.0)*0.5; L= (L +1.0)*0.5;} the     } -printf"%.4LF", ans); - fclose (stdin); - fclose (stdout); +     return 0; -}

Bzoj 3450:tyvj1952 Easy expectation/probability, dynamic planning