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  • 周武王补课

    题目叙述

    方法:

    期望DP

    #include <bita/stdc++.h>
    const int maxn = 100010;
    double p[maxn], g[maxn], f[maxn];
    int main()
    {
        int n;
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) scanf("%lf", p + i);
        for (int i = 1; i <= n; ++i)
        {
            g[i] = p[i] * (g[i - 1] + 1.f);
            f[i] = p[i] * (f[i - 1] + 2.f * g[i - 1] + 1.f) + (1.f - p[i]) * f[i - 1];
        }
        printf("%.3lf
    ", f[n]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/xmex/p/10099474.html
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