题意:
给定长度为n的非负序列x,要求计算:
答案对1e9+7取模。
数据范围:n<=5e5,0<=x[i]<=260
解法:
∑ i = 1 n ∑ j = 1 n ∑ k = 1 n ( x i & x j ) ∗ ( x j ∣ x k ) = ∑ i = 1 n ∑ j = 1 n ( x i & x j ) ∗ ∑ k = 1 n ( x j ∣ x k ) = ∑ j = 1 n ( ∑ i = 1 n ( x i & x j ) ∗ ∑ k = 1 n ( x j ∣ x k ) ) = ∑ j = 1 n f ( j ) ∗ g ( j ) 按 位 计 算 贡 献 即 可 \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n(x_{i}\&x_{j})*(x_{j}|x_{k})\\ =\sum_{i=1}^n\sum_{j=1}^n(x_{i}\&x_{j})*\sum_{k=1}^n(x_{j}|x_{k})\\ =\sum_{j=1}^n(\sum_{i=1}^n(x_{i}\&x_{j})*\sum_{k=1}^n(x_{j}|x_{k}))\\ =\sum_{j=1}^nf(j)*g(j)\\ 按位计算贡献即可 i=1∑nj=1∑nk=1∑n(xi&xj)∗(xj∣xk)=i=1∑nj=1∑n(xi&xj)∗k=1∑n(xj∣xk)=j=1∑n(i=1∑n(xi&xj)∗k=1∑n(xj∣xk))=j=1∑nf(j)∗g(j)按位计算贡献即可
code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int maxm=2e6+5;
const int mod=1e9+7;
int a[maxm];
int cnt[66];
int n;
signed main(){
ios::sync_with_stdio(0);cin.tie(0);
int T;cin>>T;
while(T--){
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];
int ans=0;
for(int j=0;j<60;j++)cnt[j]=0;
for(int i=1;i<=n;i++)for(int j=0;j<60;j++)cnt[j]+=(a[i]>>j&1);
for(int j=1;j<=n;j++){
int f=0,g=0;
for(int p=0;p<60;p++){
if(a[j]>>p&1){
f+=(1ll<<p)%mod*cnt[p]%mod;
g+=(1ll<<p)%mod*n%mod;
f%=mod,g%=mod;
}else{
g+=(1ll<<p)%mod*cnt[p]%mod;
g%=mod;
}
}
ans+=f*g%mod;
ans%=mod;
}
cout<<ans<<endl;
}
return 0;
}