bzoj1996【HNOI2010】chorus 合唱队

1996: [Hnoi2010]chorus 合唱队

Time Limit: 4 Sec Memory Limit: 64 MB

Submit: 1301 Solved: 830

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Description

Input

Output

Sample Input

4

1701 1702 1703 1704

Sample Output

8

HINT

Source

f[i][j][0]和f[i][j][1]分别表示从i到j，最后一个从前和从后加入的方案数，然后就可以DP了。(转移方程详见程序)
注意i=j和i+1=j的特殊情况。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define maxn 2005
#define mod 19650827
using namespace std;
int n;
int a[maxn],f[maxn][maxn][2];
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int main()
{
	n=read();
	F(i,1,n) a[i]=read();
	memset(f,0,sizeof(f));
	F(i,1,n) f[i][i][0]=f[i][i][1]=1;
	D(i,n-1,1) F(j,i+1,n)
	{
		if (i+1==j)
		{
			if (a[i]<a[j]) (f[i][j][0]+=1)%=mod,(f[i][j][1]+=1)%=mod;
			continue;
		}
		if (a[i]<a[i+1]) (f[i][j][0]+=f[i+1][j][0])%=mod;
		if (a[i]<a[j]) (f[i][j][0]+=f[i+1][j][1])%=mod;
		if (a[j]>a[i]) (f[i][j][1]+=f[i][j-1][0])%=mod;
		if (a[j]>a[j-1]) (f[i][j][1]+=f[i][j-1][1])%=mod;
	}
	printf("%d\n",(f[1]
[0]+f[1]
[1])%mod);
}