CodeForces - 449D Jzzhu and Numbers
Discription
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).
OutputOutput 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
很明显的套路题。
我们可以用类似FMT的方法让所有a[]的子集的cnt++,然后统计出对于每种集合 i 的f[i],即选出一堆数and的集合包含i的方案数。
显然f[i] = 2^cnt[i] - 1。
然后直接容斥就行了2333
#include<bits/stdc++.h> #define ll long long using namespace std; const int maxn=1000005,ha=1e9+7; inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;} inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;} inline int read(){ int x=0; char ch=getchar(); for(;!isdigit(ch);ch=getchar()); for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0'; return x; } int ci[maxn],n,f[maxn],ans,b[maxn]; inline void solve(){ for(int i=0;i<=20;i++) for(int j=1;j<=1e6;j++) if(ci[i]&j) f[j^ci[i]]+=f[j]; for(int i=0;i<=1e6;i++) ADD(ans,b[i]?add(ci[f[i]],ha-1):add(ha-ci[f[i]],1)); } int main(){ n=read(); ci[0]=1; for(int i=1;i<=1e6;i++) ci[i]=add(ci[i-1],ci[i-1]); b[0]=1; for(int i=1;i<=1e6;i++) b[i]=b[i^(i&-i)]^1; for(int i=1;i<=n;i++) f[read()]++; solve(); printf("%d\n",ans); return 0; }
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