Tyvj1952 Easy

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比较巧妙的是原来L^2->(l+1)^1=L^2+2*L+1这样就可以递推了

“？”的贡献及时“o”贡献的1/2。

 #include<bits/stdc++.h>
 #define LL long long
 #define LD long double
 #define  N 100005
 using namespace std;
 inline int ra()
 {
     int x=,f=; char ch=getchar();
     while (ch<'' || ch>'') {if (ch=='-') f=-; ch=getchar();}
     while (ch>='' && ch<='') {x=x*+ch-''; ch=getchar();}
     return x*f;
 }
 char s[N<<];
 double f[N<<],d[N<<];
 int main()
 {
     int n=ra();
     scanf("%s",s+);
     for (int i=; i<=n; i++)
     {
         if (s[i]=='x')
             f[i]=f[i-],d[i]=;
         else if (s[i]=='o') f[i]=f[i-]+*d[i-]+,d[i]=d[i-]+;
         else f[i]=f[i-]+d[i-]+0.5,d[i]=(d[i-]+)/;
     }
     printf("%.4lf",f[n]);
     return ;
 }