BZOJ 3450: Tyvj1952 Easy, bzojtyvj1952

BZOJ 3450: Tyvj1952 Easy, bzojtyvj1952

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 874 Solved: 646
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One day, WJMZBMR is playing osu ~~~ But he is too weak, and in some places he depends entirely on luck :(
Let's simplify the rules of this game.
There are n clicks to do. If the operation succeeds, it is o. If the operation fails, it is x. The score is calculated by comb. For a consecutive comb, there is a * a score, comb is a great continuous o.
For example, if ooxxxxooxxx is used, the score is 2*2 + 4*4 = 4 + 16 = 20.
When Sevenkplus is idle, it seems that he has played a disk. In some places, it is irrelevant to luck, either o or x, and in some places, o or x has a 50% possibility? Number.
For example, oo? Xx is a possible input.
So what is the expected osu score of WJMZBMR?
For example, oo? Xx ,? If it is o, it is oooxx => 9. If it is x, it is ooxxx => 4.
The expectation is (4 + 9)/2 = 6.5.

Input

The first line is an integer n, indicating the number of clicks.
The next string, each character is ox? One

Output

A row with a floating point number indicates the answer
Rounded to four decimal places
If you are afraid of accuracy, we recommend using long double or extended

Sample Input4
????
Sample Output4.1250


N <= 300000
Osu is fun.
WJMZBMR technology is still good (FOG), and x is rarely used.
HINT Source

We all love GYZ cups.

$ Dp [I] [1] $ indicates the expected score when the $ I $ point is reached. $ dp [I] [0] $ indicates the expected score when the $ I $ point is reached. the maximum length of $ o $ is considered for transfer for $ x $, this round cannot score $ dp [I] [1] = dp [I-1] [1], dp [I] [0] = 0 $ for $ o $, this round of score is $ (l + 1) ^ 2 = l ^ 2 + 2 * l + 1 $, $ l ^ 2 $ which we have obtained before, $2 * l + 1 $ contribution to this round of answers $? $. I thought too much at the beginning. In fact, it is very simple. It is just to divide the above two cases into two.

 1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<algorithm> 5 using namespace std; 6 const int MAXN=1e6+10; 7 const int INF=0x7fffff; 8 inline int read() 9 {10     char c=getchar();    int flag=1,x=0;11     while(c<'0'||c>'9')    {if(c=='-')    flag=-1;c=getchar();}12     while(c>='0'&&c<='9')x=x*10+c-48,c=getchar();return x*flag;13 }14 int meiyong;15 double dp[MAXN][3];16 char s[MAXN];17 int main()18 {19     meiyong=read();20     scanf("%s",s+1);21     int ls=strlen(s+1);22     for(int i=1;i<=ls;i++)23     {24         if(s[i]=='x')    dp[i][1]=dp[i-1][1],dp[i][0]=0;25         if(s[i]=='o')    dp[i][1]=dp[i-1][1]+2*dp[i-1][0]+1,dp[i][0]=dp[i-1][0]+1;26         if(s[i]=='?')    dp[i][1]=(dp[i-1][1]+dp[i-1][1]+2*dp[i-1][0]+1)/2,dp[i][0]=(dp[i-1][0]+1)/2;27     }28     printf("%.4lf",dp[ls][1]);29     return 0;30 }