cf1156F(概率dp)

弱智题一眼秒，但是挺经典的概率dp解法，所以写一篇
但是调了n久......原因是发现自己公式里有个数写错了......
不知道为什么就这点人过

直接排个序，然后很容易就能想到\(dp[i][j]\)表示抽第\(i\)轮，这轮抽到第\(j\)张牌的概率

自然就有了\(O(n^3)\)的转移

\[dp[i][j] = \sum_{k = i - 1}^{a[k]<a[j]}{\frac{1}{n-i+1}dp[i-1][k]} \]

这个打前缀和优化一下就搞定了

也有了个\(O(n^3)\)的答案

\[Ans = \sum_{i=1}^{n-1}\sum_{j=i}^{n-1}\sum_{k=j+1}^{n}\frac{1}{n-i}dp[i][j]*[a[j]==a[k]] \]

这个更简单，记个cnt数组就优化掉了一个n......
甚至不用滚动数组，512mB还是很大方的

#include <bits/stdc++.h>
#define pii pair<int,int>
#define LL long long
#define l first
#define r second
#define MAXN 100000
using namespace std;

int n;
int a[5008];
LL dp[5008][5008];
LL sum[5008][5008];
LL inv[5008];
LL cnt[5008];
LL cntsum[5008];
const LL mod = 998244353;

LL qp(LL a, LL b)
{
    LL ret = 1;
    while (b)
    {
        if (b & 1)
            ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &a[i]);
        cnt[a[i]]++;
        inv[i] = qp(i, mod - 2);
    }
    for (int i = 1; i <= n; ++i)
    {
        cntsum[i] = cntsum[i - 1] + cnt[i];
    }
    sort(a + 1, a + n + 1);
    for (int j = 1; j <= n; ++j)
    {
        dp[1][j] = inv[n];
        sum[1][j] =(sum[1][j - 1] + dp[1][j]) % mod;
    }
    for (int i = 2; i <= n; ++i)
    {
        for (int j = i; j <= n; ++j)
        {
            dp[i][j] = (sum[i - 1][cntsum[a[j] - 1]] - sum[i - 1][i - 2] + mod) % mod * inv[n - i + 1] % mod;
            sum[i][j] = (sum[i][j - 1] + dp[i][j]) % mod;
        }
    }
    LL ans = 0;
    for (int i = 1; i <= n - 1; ++i)
    {
        for (int j = i; j <= n - 1; ++j)
        {
            ans += inv[n - i] * dp[i][j] % mod * (cntsum[a[j]] - j) % mod;
            ans %= mod;
        }
    }
    printf("%I64d\n", ans * 2 % mod);
	return 0;
}