https://loj.ac/problem/2538

题目大意

略

题解

显然应该尽量先把攻击牌放完。

然后分情况讨论大力 DP 即可。

在写这题的时候被两个问题坑了好久：

1. \(k=1\)。我的计数方式没法处理这个简单问题。。。如果是在考场上的话应该多想想这样的边界情况。
2. 数组越界。开数组比较吝啬，导致访问一些下标爆范围但是没用的数据时会越界算错答案。

实现

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int rint() {
    int n, c;
    while ((c = getchar()) < '0');
    n = c - '0';
    while ((c = getchar()) >= '0') n = 10 * n + c - '0';
    return n;
}

template <int MD>
struct ModInt {
    typedef ModInt M;
    int v;
    ModInt() : v(0) {}
    ModInt(int _v) : v(_v) {}
    M& operator += (const M &r) {
        if ((v += r.v) >= MD) v -= MD;
        return *this;
    }
    M& operator *= (const M &r) {
        v = ll(v) * r.v % MD;
        return *this;
    }
    M operator + (const M &r) const { return M(*this) += r; }
    M operator * (const M &r) const { return M(*this) *= r; }
};
typedef ModInt<998244353> Mint;

const int N = 3010;
int n, m, k;
int a[N], b[N];
Mint C[2 * N][2 * N];
Mint dp[N][N], f[N][N], g[N][N];
Mint sum[N];

void first() {
    for (int i = 0; i < 2 * N; i++) {
        for (int j = 0; j <= i; j++) {
            if (j == 0) C[i][j] = Mint(1);
            else C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
        }
    }
}

void solve() {
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 0; i < n; i++) a[i] = rint();
    for (int i = 0; i < n; i++) b[i] = rint();
    sort(a, a + n, greater<int>());
    sort(b, b + n, greater<int>());

    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            dp[i][j] = Mint(0);
            f[i][j] = Mint(0);
        }
    }
    dp[0][0] = Mint(1);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= i; j++) {
            dp[i + 1][j] += dp[i][j];
            Mint x = Mint(a[i]) * dp[i][j];
            dp[i + 1][j + 1] += x;
            f[i + 1][j + 1] += x;
        }
    }

    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            dp[i][j] = Mint(0);
            g[i][j] = Mint(0);
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= i; j++) {
            dp[i + 1][j] += dp[i][j];
            Mint x = C[i][j] * Mint(b[i]) + dp[i][j];
            dp[i + 1][j + 1] += x;
            g[i + 1][j + 1] += x;
        }
    }

    if (k == 1) {
        Mint sm(0);
        for (int i = 1; i <= n; i++) {
            sm += Mint(b[i - 1]) * C[2 * n - i][m - k];
        }
        printf("%d\n", sm.v);
        return;
    }

    fill(sum, sum + n + 1, Mint(0));
    for (int i = 1; i <= n; i++) {
        for (int x = 1; x <= n; x++) {
            sum[x] += g[i][x] * C[n - i][m - k];
        }
    }

    Mint sm = sum[k];
    // 0 < t < k - 1
    for (int i = 1; i <= n; i++) {
        for (int t = 1; t < k - 1; t++) {
            sm += f[i][t] * sum[k - t];
        }
    }
    // t >= k - 1
    for (int i = 1; i <= n; i++) {
        Mint d(0);
        for (int j = 1; j <= n; j++) {
            d += Mint(b[j - 1]) * C[2 * n - i - j][m - k];
        }
        sm += d * f[i][k - 1];
    }

    printf("%d\n", sm.v);
}

int main() {
    first();
    int tc;
    scanf("%d", &tc);
    while (tc--) solve();
    return 0;
}