【BZOJ4361】isn

题目

【BZOJ4361】isn

做法

(dp_{i,j})表示以(i)结尾(j)长度，树状数组(tree_{i,j})表长度为(i)，以(<=j)结尾的个数，显然(dp_{i,j}=sumlimits_{k=1}^{pos[i]}tree[j-1][k])

从而(O(n^2logn))得到每个长度不下降子序列个数

(ans=sumlimits_{i=1}^n(g[i] imes (n-i)!-g[i+1] imes (n-i-1)! imes (i+1)))

怎么理解这句话呢？(g[i])表长度(i)的个数，我们确定了长度，自然能得到这个长度的方案数，然而有些是不合法的，因为之前已经得到不下降了不能再删

那就减去(i+1)的个数到(i)的转移

My complete code

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
typedef long long LL;
const LL maxn=3000;
const LL p=1e9+7;
inline LL Read(){
	LL x(0),f=1; char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-') f=-1; c=getchar();
	}
	while(c>='0'&&c<='9')
	    x=(x<<3)+(x<<1)+c-'0',c=getchar();
	return x*f;
}
LL n,cnt,ans;
LL tree[maxn][maxn],a[maxn],b[maxn],g[maxn],dp[maxn][maxn],pos[maxn],jc[maxn];
inline LL Lowbit(LL x){
	return x&(-x);
}
inline LL Query(LL len,LL x){
	LL ret(0);
	for(;x;x-=Lowbit(x))
	    ret=(ret+tree[len][x]);
	return ret;
}
inline void Add(LL len,LL x,LL val){
	for(;x<=n;x+=Lowbit(x))
	    tree[len][x]=(tree[len][x]+val)%p;
}
int main(){
	n=Read();
	for(LL i=1;i<=n;++i)
	    a[i]=b[i]=Read();
	sort(b+1,b+1+n);
	cnt=unique(b+1,b+1+n)-b-1;
	
	Add(0,1,1);
	for(LL i=1;i<=n;++i){
		pos[i]=lower_bound(b+1,b+1+cnt,a[i])-b;
	    for(LL j=i;j>=1;--j){
	    	dp[i][j]=Query(j-1,pos[i])%p;
	    	Add(j,pos[i],dp[i][j]);
		}
	}
	
	for(LL i=1;i<=n;++i)
	    for(LL j=1;j<=n;++j)
	        g[i]=(g[i]+dp[j][i])%p;
	jc[1]=1;
	for(LL i=2;i<=n;++i)
	    jc[i]=jc[i-1]*i%p;
	for(LL i=1;i<=n;++i)
	    ans=(ans+g[i]*jc[n-i]%p-g[i+1]*jc[n-i-1]%p*(i+1)%p+p)%p;
	printf("%lld
",ans);
    return 0;
}