loj2538. 「PKUWC2018」Slay the Spire

Posted 2022-09-20 psimonw

题意

略。

题解

考虑到尽可能多选强化卡是更优的，所以如果可以，最后只要选最大的一张攻击即可（除非强化卡不够了）。
那么按照这个思路，先把两个序列从大到小排序。
记录\(f_i, j\)表示选了\(i\)张强化卡，其中最后一张是第\(j\)张的所有方案的强化倍数的和。
则有
\[ f_i, j = a_i \sum_k = 1 ^ j - 1 f_i - 1, k \]
对于攻击卡也类似。
但这并不是我们想要的。
我们想要的\(F_i, j\)应该是分到\(i\)张强化卡，（最优地）最后打出了\(j\)张强化卡的期望强化倍数之和。
因此，计算\(F_p, q\)的式子应该是
\[ F_p, q = \sum_i = 1 ^ n f_i, q \binomn - ip - q \]
单组复杂度\(\mathcal O(n ^ 2 + nm)\)。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3005, mod = 998244353;
int T, n, m, k, ans;
int a[N], b[N], s[N], c[N][N], f[N][N], g[N][N];
void prepare () 
    c[0][0] = 1;
    for (int i = 1; i < N; ++i) 
        c[i][0] = 1;
        for (int j = 1; j <= i; ++j) 
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
        
    

int F (int p, int q) 
    if (p < q) 
        return 0;
    
    if (!q) 
        return c[n][p];
    
    int ret = 0;
    for (int i = 1; i <= n; ++i) 
        ret = (1ll * f[q][i] * c[n - i][p - q] + ret) % mod;
    
    return ret;

int G (int p, int q) 
    if (p < q) 
        return 0;
    
    int ret = 0;
    for (int i = 1; i <= n; ++i) 
        ret = (1ll * g[q][i] * c[n - i][p - q] + ret) % mod;
    
    return ret;

int main () 
    prepare();
    for (scanf("%d", &T); T; --T) 
        scanf("%d%d%d", &n, &m, &k);
        for (int i = 1; i <= n; ++i) 
            scanf("%d", &a[i]);
        
        for (int i = 1; i <= n; ++i) 
            scanf("%d", &b[i]);
        
        sort(a + 1, a + n + 1), reverse(a + 1, a + n + 1);
        sort(b + 1, b + n + 1), reverse(b + 1, b + n + 1);
        for (int i = 1; i <= n; ++i) 
            f[1][i] = a[i];
            s[i] = (s[i - 1] + a[i]) % mod;
        
        for (int i = 2; i <= n; ++i) 
            for (int j = 1; j <= n; ++j) 
                f[i][j] = 1ll * a[j] * s[j - 1] % mod;
            
            for (int j = 1; j <= n; ++j) 
                s[j] = (s[j - 1] + f[i][j]) % mod;
            
        
        for (int i = 1; i <= n; ++i) 
            g[1][i] = b[i];
            s[i] = (s[i - 1] + b[i]) % mod;
        
        for (int i = 2; i <= n; ++i) 
            for (int j = 1; j <= n; ++j) 
                g[i][j] = (1ll * b[j] * c[j - 1][i - 1] + s[j - 1] + mod) % mod;
            
            for (int j = 1; j <= n; ++j) 
                s[j] = (s[j - 1] + g[i][j]) % mod;
            
        
        ans = 0;
        for (int i = 0; i < m; ++i) 
            if (i < k) 
                ans = (1ll * F(i, i) * G(m - i, k - i) + ans) % mod;
             else 
                ans = (1ll * F(i, k - 1) * G(m - i, 1) + ans) % mod;
            
        
        printf("%d\n", ans);
    
    return 0;