2020-2021 ACM-ICPC Brazil Subregional Programming Contest A. Sticker Album

期望dp

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https://codeforces.com/gym/102861/problem/A

大小为n的背包,无限个物品随机在$[a,b]$之间,计算E(n),E(n)表示随机选择物品,填满背包需要的期望物品数量

tip1: 普通期望dp是n*(b-a+1)的复杂度, 观察dp公式可以发现可以用滑块保存下一个状态

• $dp[i]=\frac{1}{l}\ast dp[i-a]+\frac{1}{l}\ast dp[i-a+1]+\cdots +\frac{1}{l}\ast dp[i-b]+1$

• $sum=dp[i-a]+dp[i-a+1]+\cdots +dp[i-b]$

tip2: a = 0 的时候存在dp[i] -> dp[i-1]的转移需要特殊考虑

• $dp[i]=\frac{1}{l}\ast dp[i]+\frac{1}{l}\ast dp[i+1]+\cdots +\frac{1}{l}\ast dp[i-b]+1$

                    $\Downarrow$

  $\frac{l-1}{l}\ast dp[i]=\frac{1}{l}\ast dp[i+1]+\cdots +\frac{1}{l}\ast dp[i-b]+1$

• $sum=dp[i+1]+\cdots +dp[i-b]$

    ll n, a, b, len;
    int main()
    {
        cin >> n >> a >> b;
        len = b - a + 1;
        for (int i = n - 1; i >= 0; i--) {
            if (a)
                dp[i] = (sum / len + 1.0);
            else
                dp[i] = (sum + len) / (len - 1);
            sum -= dp[i + b];
            if (a)
                sum += dp[i + a - 1];
            else
                sum += dp[i];
        }
        cout << fixed << setprecision(10) << dp[0] << endl;
        return 0;
    }