[codeforces 734F]Anton and School

【题目大意】给出长度为n(<=200000)的两个序列b和c，bc与a的关系如下图，求a

【解题思路】因为(a and b)+(a or b)=a+b，所以有Σa=(Σb+Σc)/(2*n)，然后就可以通过a[i]=(b[i]+c[i]-Σa)/n求解

【呆马】

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<iostream>
using namespace std;
const int N=200001;
long long n,i,sum,B,C,j,k,a
,b
,c
,d
,s[30],bit[30];
void ns(){printf("-1"); exit(0);}
int main()
{
scanf("%I64d\n",&n);
for (i=1;i<=n;i++) scanf("%I64d",&b[i]);
for (scanf("\n"),i=1;i<=n;i++) scanf("%I64d",&c[i]),d[i]=b[i]+c[i],sum+=d[i];
if (sum%(2*n)) ns();
sum/=2*n;
for (i=1;i<=n;i++)
{
a[i]=d[i]-sum;
if (a[i]%n) ns();
a[i]/=n;
}
for (s[0]=i=1;i<=29;i++) s[i]=s[i-1]<<1;
for (i=1;i<=n;i++)
for (j=a[i],k=0;k<=29;k++,j>>=1) bit[k]+=j&1;
for (i=1;i<=n;i++)
{
B=C=0;
for (j=a[i],k=0;k<=29;k++,j>>=1)
if (j&1) B+=s[k]*bit[k],C+=s[k]*n;
else C+=s[k]*bit[k];
if (B!=b[i] || C!=c[i]) ns();
}
for (i=1;i<=n;i++) printf("%I64d ",a[i]);
}