A Expectation (Unaccepted)

题目大意：

咕咕

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B Sequence (Unaccepted)

题目大意：

咕咕

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C Counting

题目大意：给出n个点，加E条不同的有向边，没加一条记录下强联通分量个数，组成一个序列，问有多少种这样的序列

考虑最优策略一定是 1,2,...,i 连一个大环，后面全是单独的强连通分量，那么可行的边数不超过 frac{n(n-1)+i(i-1)}{2}

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#include <cstdio>

using namespace std;
const int N = 105, M = 998244353;

int f[N*N][N][N], s[N*N][N][N], n;
//f[e][i][j] : 用了e条边，1所在的scc大小为i，反向连了j条边的方案数
//s[e][i][j] : f[e][1][j] - f[e][i][j] 的和

int main() {
    scanf("%d", &n); f[0][1][0] = 1;
    for (int i = 1; i <= n; ++i) s[0][i][0] = 1;
    int lim = n * (n - 1);
    for (int e = 1; e <= lim; ++e) {
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (e >= i + j - 1 && e <= i * (i - 1) + (n - i) * (n - i - 1) / 2 + i * (n - i))
                    f[e][i][j] = (f[e-1][i][j] + s[e-1][i-1][j-1]) % M;
                s[e][i][j] = (s[e][i-1][j] + f[e][i][j]) % M;
            }
        }
    }
    for (int i = 1; i <= lim; ++i) {
        int ans = 0;
        for (int j = 0; j < n; ++j)
            if ((ans += s[i][n][j]) >= M) ans -= M;
        printf("%d ", ans);
    }
    return 0;
}