sosdp 二进制子集的贡献

可以O(n*2^n)算出n位每个mask值所包含子集的二进制码下标的贡献

比如f[5]=a[0]+a[1]+a[4]+a[5]这种的，101包含了000,001,100,101

#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<stack>
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
#include<stdint.h>

using namespace std;         //

#define ll long long  
#define ull unsigned long long
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){
    ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{int x;int y;};
inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
void ex(){puts("NO");exit(0);}
const int dx[8]={0,0,-1,1,-1,1,-1,1};
const int dy[8]={1,-1,0,0,-1,1,1,-1};
const int inf=0x3f3f3f3f; 
const ll Linf=0x3f3f3f3f3f3f3f3fLL;
const ll Mod=1e18+7;
const double eps=1e-6;
const double pi=acos(-1.0);

const int maxn=1e6+33;

int n;
int a[maxn],f[maxn];

int main(){
    scanf("%d",&n);
    for(int i=0;i<(1<<n);i++){
        scanf("%d",&a[i]);
    }
    for(int i=0;i<(1<<n);i++){
        f[i]=a[i];
    }
    for(int i=0;i<n;i++){
        for(int msk=0;msk<(1<<n);msk++){
            if(msk & (1<<i))
                f[msk]+=f[msk^(1<<i)];
        }
    }
}

来源是cf上的一篇博客

核心思想就是从低位枚举到高位，

f[mask][i]表示mask码低i位子集的贡献

如果mask的第i位是1，那么f[mask][i]=f[mask][i-1]+f[mask^(1<<i)][i-1]，很巧妙地以第i位的01作为子集划分

如果mask的第i位是0，那么f[mask][i]=f[mask][i-1]

------------------------------------------------------------------------------------------------------------------------

例题SPECIAL PAIRS

1e5的数列问有多少个pair满足a[i]&a[j]=0

对于a[i]，与他匹配的a[j]一定都贡献进a[i]的补集，那么答案就是sigma(f[补ai])

#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<stack>
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
#include<stdint.h>

using namespace std;         //

#define ll long long  
#define ull unsigned long long
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){
    ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{int x;int y;};
inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
void ex(){puts("NO");exit(0);}
const int dx[8]={0,0,-1,1,-1,1,-1,1};
const int dy[8]={1,-1,0,0,-1,1,1,-1};
const int inf=0x3f3f3f3f; 
const ll Linf=0x3f3f3f3f3f3f3f3fLL;
const ll Mod=1e18+7;
const double eps=1e-6;
const double pi=acos(-1.0);

const int maxn=1e6+3;

int n;int b[100011];
int a[maxn],f[maxn];
int T;
int main(){
    read(T);
    while(T--){
        memset(a,0,sizeof a);
        read(n);
        for(int i=0;i<n;i++){
            read(b[i]);
            a[b[i]]++;
        }

        for(int i=0;i<maxn;i++){
            f[i]=a[i];
        }
        int lim=log2(maxn);
        for(int i=0;i<lim;i++){
            for(int msk=0;msk<maxn;msk++){
                if(msk & (1<<i))
                    f[msk]+=f[msk^(1<<i)];
            }
        }
        ll ans=0;

        for(int i=0;i<n;i++){
            ans+=f[((1<<lim)-1) ^ b[i]];
        }
        printf("%lld\n",ans);
    }
}