BZOJ 5161: 最长上升子序列 状压dp+查分

好神啊 ~ 

打表程序：

#include <cstdio> 
#include <cstring> 
#include <algorithm>  
#define N 140000000
#define ll long long  
#define mod 998244353 
#define setIO(s) freopen(s".in","r",stdin) , freopen(s".out","w",stdout)
using namespace std;        
int f[2][N]; 
int cnt[N];    
ll qpow(ll x,ll y) { 
    ll ans=1; 
    while(y) {
        if(y&1) {
            ans=ans*x%mod;  
        }
        x=x*x%mod; 
        y>>=1; 
    }
    return ans; 
}   
int main() {           
    // setIO("input"); 
    for(int n=1;n<=30;n+=2) {          
        int i,j,k,d,t,pos;  
        ll ans=0,fac=1; 
        --n;         
        f[0][0]=1; 
        for(d=i=1;i<=n;++i,d^=1) {
            memset(f[d],0,sizeof(int)*(1<<i));   
            for(j=0;j<(1<<(i-1));++j) {     
                f[d][j<<1]=(f[d][j<<1]+f[d^1][j])%mod,pos=-1;    
                for(k=i-1;k>=0;--k) {
                    t=((j>>k)<<(k+1))|(1<<k)|(j&((1<<k)-1));     
                    if(j&(1<<k)) pos=k;        // 这里有一   
                    if(pos>=0) t^=(1<<(pos+1));     
                    f[d][t]=(f[d][t]+f[d^1][j])%mod;   
                }
            }
        }
        for(i=1;i<(1<<n);++i) {
            cnt[i]=cnt[i-(i&-i)]+1;   
        }
        for(i=0;i<(1<<n);++i) {
            ans=(ans+1ll*f[n&1][i]*(cnt[i]+1))%mod;     
        }
        for(i=1;i<=n+1;++i)   {
            fac=fac*i%mod;  
        }
        printf("%lld
",ans*qpow(fac,mod-2)%mod);  
    }
    return 0; 
}

　　

表： 

#include <cstdio>
int ans[40]={
    1,
    499122178,
    2,
    915057326,
    540715694,
    946945688,
    422867403,
    451091574,
    317868537,
    200489273,
    976705134,
    705376344,
    662845575,
    331522185,
    228644314,
    262819964,
    686801362,
    495111839,
    947040129,
    414835038,
    696340671,
    749077581,
	301075008,
	314644758,
	102117126,
	819818153,    
	273498600,
	267588741,
};  
int main() {
    int n; 
    scanf("%d",&n);  
	printf("%d
",ans[n-1]);  
	return 0; 
}