洛谷P1365 WJMZBMR打osu! / Easy 题解 期望DP

题目链接：https://www.luogu.com.cn/problem/P1365

题解地址：https://www.luogu.com.cn/blog/five20/solution-p1365

实现代码如下：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 300030;
int n;
char s[maxn];
double f[maxn], g[maxn];
int main() {
    scanf("%d%s", &n, s+1);
    for (int i = 1; i <= n; i ++) {
        if (s[i] == 'x') f[i] = f[i-1], g[i] = 0;
        else if (s[i] == 'o') f[i] = f[i-1] + 2 * g[i-1] + 1, g[i] = g[i-1] + 1;
        else f[i] = f[i-1] + g[i-1] + 0.5, g[i] = (g[i-1] + 1) / 2;
    }
    printf("%.4lf\n", f[n]);
    return 0;
}

洛谷P1365 WJMZBMR打osu! / Easy 题解 期望DP

原文：https://www.cnblogs.com/quanjun/p/12261551.html