codeforces 449D - Jzzhu and Numbers 高维前缀和 容斥

D. Jzzhu and Numbers

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Jzzhu have n non-negative integers a1, a2, ..., an.
We will call a sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n) a
group of size k.

Jzzhu wonders, how many groups exists such that ai1 & ai2 &
... & aik = 0 (1 ≤ k ≤ n)?
Help him and print this number modulo 1000000007 (109 + 7).
Operation x & y denotes
bitwise AND operation of two numbers.

Input

The first line contains a single integer n (1 ≤ n ≤ 106).
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 106).

Output

Output a single integer representing the number of required groups modulo 1000000007 (109 + 7).

Examples

input

3
2 3 3

output

0

input

4
0 1 2 3

output

10

input

6
5 2 0 5 2 1

output

53

这不是裸的高维前缀和吗。。。

容斥搞一搞就好了A。。。

不过想一想。BJ就是为了练一下高维前缀和查的题

好像不会高维前缀和的话没法做呢。。。

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<set>
#include<map>
using namespace std;

typedef long long ll;

inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
void print(int x)
{if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}

const int N=1001000,mod=int(1e9)+7;

int f[N<<1];

inline int bit(int x)
{
int res=0;
while(x)
{
res^=(x&1);
x>>=1;
}
return res;
}

inline int qpow(int x,int y)
{
int res=1;
while(y)
{
if(y&1) res=1ll*res*x%mod;
x=1ll*x*x%mod;
y>>=1;
}
return res;
}

int main()
{
int n=read();
register int i,j;
for(i=1;i<=n;++i) f[read()]++;
for(j=0;j<21;++j)
for(i=1<<20;i;i--)
if((i>>j)&1)
(f[i^(1<<j)]+=f[i])%=mod;
int ans=0;
for(i=0;i<(1<<20);++i)
{
bit(i) ? ans-=qpow(2,f[i]) : ans+=qpow(2,f[i]);
ans%=mod;
}
cout<<(ans+mod)%mod<<endl;
return 0;
}