51nod1407 与与与与

Posted 2022-02-04 ivorysi

【51nod】1407 与与与与

设\(f(x)\) 为\(A_i \& x == x\)的\(A_i\)的个数

设\(g(x)\)为\(x\)里1的个数

\(\sum_i = 0^2^20 (-1)^g(x)2^f(x)\)

\(f(x)\)就是按位取反之后的一个FMT卷积，把判断条件改成这一位不存在即可

也可以用FWT的与卷积直接卷起来

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) 
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') 
        if(c == '-') f = -1;
        c = getchar();
    
    while(c >= '0' && c <= '9') 
        res = res * 10 +c - '0';
        c = getchar();
    
    res *= f;

template<class T>
void out(T x) 
    if(x < 0) x = -x;putchar('-');
    if(x >= 10) 
        out(x / 10);
    
    putchar('0' + x % 10);

const int MOD = 1000000007;
int N;
int a[(1 << 20) + 5],cnt[(1 << 20) + 5],pw[1000005];
int lowbit(int x) 
    return x & (-x);

int inc(int a,int b) 
    return a + b >= MOD ? a + b - MOD : a + b;

int mul(int a,int b) 
    return 1LL * a * b % MOD;

void update(int &x,int y) 
    x = inc(x,y);

void Solve() 
    memset(a,0,sizeof(a));
    int d;
    pw[0] = 1;
    for(int i = 1 ; i <= N ; ++i) 
        read(d);a[d]++;
        pw[i] = mul(pw[i - 1],2);
    
    for(int j = 0 ; j < 20 ; ++j) 
        for(int S = (1 << 20) - 1 ; S >= 0 ; --S) 
            if(!((S >> j) & 1)) 
                a[S] += a[S ^ (1 << j)];
            
        
    
    int ans = 0;
    for(int S = 0 ; S < (1 << 20) ; ++S) 
        if(S) cnt[S] = cnt[S - lowbit(S)] + 1;
        int t;
        if(cnt[S] & 1) t = MOD - 1;
        else t = 1;
        update(ans,mul(t,pw[a[S]]));
    
    out(ans);enter;

int main()
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    while(scanf("%d",&N) != EOF) Solve();