题目大意:一开始有 $2n$ 张牌,每张牌上都写着一个数字 $w_i$,有两种的牌,每种类型各 $n$ 张:
1.攻击牌:打出后对对方造成牌上的数字的伤害。
2.强化牌:打出后,设数字为 $x$,则其他剩下的攻击牌的数字都会乘上 $x$。保证强化牌上的数字都大于$1$。
现在会等概率随机从卡组中抽出 $m$ 张牌,但最多打出 $k$ 张牌,假设会采取能造成最多伤害的策略,求她期望造成多少伤害。
题解:发现有若可以的话,先从大到小打强化牌,再从大到小打攻击牌(至少打一张),这样会造成最大伤害,然后我就不会了。。。
卡点:无
C++ Code:
#include <cstdio> #include <algorithm> #define maxn 3010 #define int long long using namespace std; const int mod = 998244353; int Tim, n, m, k; int a[maxn], b[maxn]; int fac[maxn], inv[maxn]; int f[maxn][maxn], g[maxn][maxn]; int sf[maxn][maxn], sg[maxn][maxn]; inline bool cmp(int a, int b) {return a > b;} int C(int a, int b) { if (b > a) return 0; return fac[a] * inv[b] % mod * inv[a - b] % mod; } int F(int a, int b) { if (a < b) return 0; int ans = 0; for (int i = 0; i <= n; i++) ans = (ans + f[b][i] * C(n - i, a - b)) % mod; return ans; } int G(int a, int b) { if (a < b) return 0; int ans = 0; for (int i = 0; i <= n; i++) ans = (ans + g[b][i] * C(n - i, a - b)) % mod; return ans; } signed main() { fac[0] = fac[1] = inv[0] = inv[1] = 1; for (int i = 2; i <= maxn; i++) { fac[i] = fac[i - 1] * i % mod; inv[i] = inv[mod % i] * (mod - mod / i) % mod; } for (int i = 2; i <= maxn; i++) inv[i] = inv[i] * inv[i - 1] % mod; scanf("%lld", &Tim); while (Tim --> 0) { scanf("%lld%lld%lld", &n, &m, &k); for (int i = 1; i <= n; i++) scanf("%lld", &a[i]); sort(a + 1, a + n + 1, cmp); f[0][0] = 1; for (int i = 0; i <= n; i++) sf[0][i] = 1; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { f[i][j] = a[j] * sf[i - 1][j - 1] % mod; sf[i][j] = (sf[i][j - 1] + f[i][j]) % mod; } } for (int i = 1; i <= n; i++) scanf("%lld", &b[i]); sort(b + 1, b + n + 1, cmp); for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { g[i][j] = (b[j] * C(j - 1, i - 1) + sg[i - 1][j - 1]) % mod; sg[i][j] = (sg[i][j - 1] + g[i][j]) % mod; } } int ans = 0; for (int i = 0; i <= m; i++) { if (i < k) ans = (ans + F(i, i) * G(m - i, k - i)) % mod; else ans = (ans + F(i, k - 1) * G(m - i, 1)) % mod; } printf("%lld ", ans); } return 0; }