Jzzhu and Numbers

最新推荐文章于 2021-01-18 13:59:15 发布

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最新推荐文章于 2021-01-18 13:59:15 发布

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原文链接：http://www.cnblogs.com/dyzll/p/6536551.html

本文探讨了一种涉及位运算和组合计数的问题，通过容斥原理解决特定数学挑战。介绍了一个高效的算法来计算满足特定条件的整数组数量，并提供了详细的代码实现。

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 modulo1000000007 (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

input

4
0 1 2 3

output

input

6
5 2 0 5 2 1

output

53
分析：暴力肯定不行了，考虑容斥；
　　　答案为ans=Σ(-1)^f(x)*pow(2,num[x])，f(x)代表x中二进制1的个数，num[x]代表a[k]&x=x的个数；
　　　求num[x]=Σcnt[k](a[k]&x=x)，即为高维前缀和，与spoj tle类似；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e6+10;
const int N=2e2+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t,num[1<<20];
ll dp[1<<20],p[maxn];
int main()
{
    int i,j;
    p[0]=1;
    rep(i,1,maxn-10)p[i]=p[i-1]*2%mod;
    scanf("%d",&n);
    rep(i,1,n)scanf("%d",&j),dp[j]++;
    rep(i,0,19)
    {
        rep(j,0,(1<<20)-1)
        {
            if((~j)&(1<<i))(dp[j]+=dp[j^(1<<i)])%=mod;
            if(j&(1<<i))num[j]++;
        }
    }
    ll ret=0;
    rep(i,0,(1<<20)-1)
    {
        if(num[i]&1)ret=(ret-p[dp[i]]+mod)%mod;
        else ret=(ret+p[dp[i]])%mod;
    }
    printf("%lld\n",ret);
    return 0;
}