CodeForces 734F Anton and School

位运算。

两个数的和：$A+B=(AandB)+(AorB)$，那么$b[i]+c[i]=n*a[i]+suma$。可以解出一组解，然后再按位统计贡献验证一下。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0);
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = ;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c))
    {
        x = x *  + c - '';
        c = getchar();
    }
}

LL sum,sumc,sumb;
LL t[],g[],a[],b[],c[];
int n;

int main()
{
    g[]=; for(int i=;i<=;i++) g[i]=g[i-]*;
    cin>>n;
    for(int i=;i<=n;i++)
    {
        cin>>b[i];
        sumb=sumb+b[i];
    }

    for(int i=;i<=n;i++)
    {
        cin>>c[i];
        sumc=sumc+c[i];
    }

    if((sumb+sumc)%(LL)(*n)!=)
    {
        printf("-1\n");
        return ;
    }

    sum=(sumb+sumc)/(LL)(*n);

    for(int i=;i<=n;i++)
    {
        if((b[i]+c[i]-sum)<)
        {
            printf("-1\n");
            return ;
        }

        if((b[i]+c[i]-sum)%(LL)n!=)
        {
            printf("-1\n");
            return ;
        }

        a[i]=(b[i]+c[i]-sum)/(LL)n;
    }

    for(int i=;i<=n;i++)
        for(int j=;j<=;j++) if(a[i]&g[j]) t[j]++;

    bool fail=;

    for(int i=;i<=n;i++)
    {
        long long sum=;
        for(int j=;j<=;j++) if(a[i]&g[j]) sum=sum+t[j]*g[j];
        if(sum!=b[i]) fail=;

        sum=;
        for(int j=;j<=;j++)
        {
            if(a[i]&g[j]) sum=sum+n*g[j];
            else sum=sum+t[j]*g[j];
        }
        if(sum!=c[i]) fail=;
    }

    if(fail==)
    {
        for(int i=;i<=n;i++) cout<<a[i]<<" ";
        cout<<endl;
    }
    else cout<<"-1"<<endl;

    return ;
}