题目链接
题解
既已开题
那就已经没有什么好害怕的了
由题目中奇怪的条件我们可以特判掉(n - k)为奇数时答案为(0)
否则我们要求的就是糖果大于药片恰好有(frac{n - k}{2} + k)个的方案数，我们记为(K)
思路1##
直接求恰好不好求，想到二项式反演：
如果有
[b_k = sumlimits_{i = k}^{n} {i choose k} a_i
]
那么有
[a_k = sumlimits_{i = k}^{n} (-1)^{i - k} {i choose k} b_i
]
如果我们令(a_k)为恰好有(K)对糖果大于药片的方案数
从式子来看，对于每一种(a_i)的情况，(b_k)都从中选出任意(k)个
那么(b_k)就是任选(k)个满足条件，剩余随便选的方案数
问题就转化为了求出(b_k)，显然可以(dp)
我们先将糖果与药片分别升序排序
我们设(f[i][j])表示前(i)个糖果选出(j)个和药片匹配，都大于药片的方案数
我们记(pos_i)为第(i)个糖果比药片大的最大的位置，那么如果第(i)个糖果参与匹配，就有(pos_i)种选择
由于糖果是升序排序，所以剩余(j - 1)个糖果选择范围一定不比(i)大，所以(i)实际剩余(pos_i - (j - 1))种选择
那么有
[f[i][j] = f[i - 1][j] + f[i - 1][j - 1] * (pos_i - j + 1)
]
求出(f[n][i])，则(b_k = f[n][k] * (n - k)!)
然后再用二项式反演即可求出(a_k)
思路2##
如果你不知道二项式反演，其实有一个更直观的方法
同样是求出(f[n][i])后求出(b_i)
我们令(ans[k])表示恰好(k)个糖果大于药片的方案数
显然
[ans[n] = b_n
]
当(k < n)时，(b_k)多算了恰好为(k + 1)、(k + 2)、(k + 3......)的方案数，减去即可
[ans[k] = b_k - sumlimits_{i = k + 1}^{n} ans[i]
]
这样也是对的
PS##
由思路2其实我们可以看到二项式反演实际上就是一个容斥的结果
思路一
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 2005,maxm = 100005,INF = 1000000000,P = 1000000009;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
LL f[maxn][maxn],C[maxn][maxn],fac[maxn];
int a[maxn],b[maxn];
int n,K;
void init(){
fac[0] = 1;
for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % P;
for (int i = 0; i <= n; i++){
C[i][0] = C[i][i] = 1;
for (int j = 1; j <= (i >> 1); j++)
C[i][j] = C[i][i - j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
}
sort(a + 1,a + 1 + n);
sort(b + 1,b + 1 + n);
}
int main(){
n = read(); K = read();
if ((n - K) & 1) {puts("0"); return 0;}
K = ((n - K) >> 1) + K;
for (int i = 1; i <= n; i++) a[i] = read();
for (int i = 1; i <= n; i++) b[i] = read();
init();
f[0][0] = 1;
for (int i = 1; i <= n; i++){
f[i][0] = f[i - 1][0];
int pos = n;
while (pos && b[pos] >= a[i]) pos--;
for (int j = 1; j <= i; j++)
f[i][j] = (f[i - 1][j] + f[i - 1][j - 1] * max(pos - j + 1,0) % P) % P;
}
LL ans = 0,t;
for (int i = K; i <= n; i++){
t = ((i - K) & 1) ? -1 : 1;
ans = ((ans + t * C[i][K] % P * (f[n][i] * fac[n - i] % P) % P) % P + P) % P;
}
printf("%lld
",ans);
return 0;
}
思路2
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 2005,maxm = 100005,INF = 1000000000,P = 1000000009;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
LL f[maxn][maxn],C[maxn][maxn],fac[maxn],ans[maxn];
int a[maxn],b[maxn];
int n,K;
void init(){
fac[0] = 1;
for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % P;
for (int i = 0; i <= n; i++){
C[i][0] = C[i][i] = 1;
for (int j = 1; j <= (i >> 1); j++)
C[i][j] = C[i][i - j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
}
sort(a + 1,a + 1 + n);
sort(b + 1,b + 1 + n);
}
int main(){
n = read(); K = read();
if ((n - K) & 1) {puts("0"); return 0;}
K = ((n - K) >> 1) + K;
for (int i = 1; i <= n; i++) a[i] = read();
for (int i = 1; i <= n; i++) b[i] = read();
init();
f[0][0] = 1;
for (int i = 1; i <= n; i++){
f[i][0] = f[i - 1][0];
int pos = n;
while (pos && b[pos] >= a[i]) pos--;
for (int j = 1; j <= i; j++)
f[i][j] = (f[i - 1][j] + f[i - 1][j - 1] * max(pos - j + 1,0) % P) % P;
}
ans[n] = f[n][n];
for (int i = n - 1; i >= K; i--){
ans[i] = f[n][i] * fac[n - i] % P;
for (int j = i + 1; j <= n; j++)
ans[i] = (ans[i] - C[j][i] * ans[j] % P) % P;
ans[i] = (ans[i] + P) % P;
}
printf("%lld
",ans[K]);
return 0;
}