3450: Tyvj1952 Easy 概率与期望 DP

维护一个最长连续’o’序列的期望就好了。。

#include<iostream>#include<cstdio>using namespace std;int n;char s[300005];double p1[300005],p2[300005];double ans;int main(){    scanf("%d",&n);    scanf("%s",s+1);    for (int i=1;i<=n;i++)    {        double p;        switch (s[i])        {            case 'o':p=1.0; break;            case 'x':p=0.0; break;            case '?':p=0.5; break;        }        p1[i]=(p1[i-1]+1)*p;        p2[i]=(p2[i-1]+2*p1[i-1]+1)*p;        ans+=p2[i-1]*(1-p);    }    ans+=p2[n];    printf("%.4lf",ans);    return 0;}