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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。
1 #include<bits/stdc++.h> 2 #define LL long long 3 #define LD long double 4 #define N 100005 5 using namespace std; 6 inline int ra() 7 { 8 int x=0,f=1; char ch=getchar(); 9 while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();} 10 while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();} 11 return x*f; 12 } 13 char s[N<<2]; 14 double f[N<<2],d[N<<2]; 15 int main() 16 { 17 int n=ra(); 18 scanf("%s",s+1); 19 for (int i=1; i<=n; i++) 20 { 21 if (s[i]=='x') 22 f[i]=f[i-1],d[i]=0; 23 else if (s[i]=='o') f[i]=f[i-1]+2*d[i-1]+1,d[i]=d[i-1]+1; 24 else f[i]=f[i-1]+d[i-1]+0.5,d[i]=(d[i-1]+1)/2; 25 } 26 printf("%.4lf",f[n]); 27 return 0; 28 }