「PKUWC2018」Slay the Spire

题面

题解

$ ext{PKUWC2018}$就是$mod;998244353$场。。。

首先可以发现，应该先打完强化牌后再打攻击牌，牌都尽量打大的。

所以，打$K−1$张强化，$1$ 张攻击是最优的。当然如果强化用完了就打攻击。

设$F(x,y)$表示抽出$x$张强化，前$y$张的乘积和

类似的设$G(x,y)$

令$f(i,j)$表示用$i$张强化，最小值为$j$的贡献

类似的设$g(x,y)$

$ ext{O}(n^2)$求出以上函数即可。

（不是我说，真的恶心）

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x));

namespace IO
{
	const int BUFSIZE = 1 << 20;
	char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;
	inline char getchar() { if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin); return *is++; }
}

inline int read()
{
	int data = 0, w = 1;
	char ch = IO::getchar();
	while(ch != '-' && (ch < '0' || ch > '9')) ch = IO::getchar();
	if(ch == '-') w = -1, ch = IO::getchar();
	while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::getchar();
	return data*w;
}

const int maxn(3010), Mod(998244353);
int T, n, m, K, a[maxn], b[maxn], C[maxn][maxn];
int f[maxn][maxn], g[maxn][maxn], sum[maxn][maxn];

int F(int x, int y)
{
	if(x < y) return 0;
	if(!y) return C[n][x];
	int ans = 0;
	for(RG int i = x - y + 1; i <= n - y + 1; i++)
		ans = (ans + 1ll * C[i - 1][x - y] * f[y][i] % Mod) % Mod;
	return ans;
}

int G(int x, int y)
{
	if(x < y) return 0;
	int ans = 0;
	for(RG int i = x - y + 1; i <= n - y + 1; i++)
		ans = (ans + 1ll * C[i - 1][x - y] * g[y][i] % Mod) % Mod;
	return ans;
}

int main()
{
#ifndef ONLINE_JUDGE
	file(cpp);
#endif
	for(RG int i = 0; i <= 3000; i++)
	{
		C[i][0] = 1;
		for(RG int j = 1; j <= i; j++)
			C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % Mod;
	}
	T = read();
	while(T--)
	{
		n = read(), m = read(), K = read();
		for(RG int i = 1; i <= n; i++) a[i] = read();
		for(RG int i = 1; i <= n; i++) b[i] = read();
		std::sort(a + 1, a + n + 1);
		std::sort(b + 1, b + n + 1);
		for(RG int i = 0; i <= n; i++)
			for(RG int j = 0; j <= n; j++) f[i][j] = g[i][j] = 0;
		for(RG int i = 1; i <= n; i++)
			f[1][i] = a[i], sum[1][i] = (sum[1][i - 1] + a[i]) % Mod;
		for(RG int i = 2; i <= n; i++)
		{
			for(RG int j = 1; j <= n - i + 1; j++)
				f[i][j] = 1ll * (sum[i - 1][n] -
					sum[i - 1][j] + Mod) % Mod * a[j] % Mod;
			for(RG int j = 1; j <= n; j++)
				sum[i][j] = (sum[i][j - 1] + f[i][j]) % Mod;
		}
		for(RG int i = 1; i <= n; i++)
			g[1][i] = b[i], sum[1][i] = (sum[1][i - 1] + b[i]) % Mod;
		for(RG int i = 2; i <= n; i++)
		{
			for(RG int j = 1; j <= n - i + 1; j++)
				g[i][j] = (1ll * (sum[i - 1][n] - sum[i - 1][j] + Mod) % Mod
						+ 1ll * b[j] * C[n - j][i - 1] % Mod) % Mod;
			for(RG int j = 1; j <= n; j++)
				sum[i][j] = (sum[i][j - 1] + g[i][j]) % Mod;
		}

		int ans = 0;
		for(RG int i = 0; i < m; i++)
			if(K - 1 <= i) ans = (ans + 1ll * F(i, K - 1)
					* G(m - i, 1) % Mod) % Mod;
			else ans = (ans + 1ll * F(i, i) * G(m - i, K - i) % Mod) % Mod;
		printf("%d
", ans);
	}
	return 0;
}