思路：转自别人的题解：
分析：只要知道a[i]+a[j]=a[i]|a[j]+a[i]&a[j]就好做了，我们令sum=∑ni=1a[i]，然后我们会得到c[i]=∑nj=1(a[i]+a[j])−b[i]=n∗a[i]+sum−b[i]即b[i]+c[i]=n∗a[i]+sum然后就可以求出a数组啦，最后我们还要检查一下a数组是否能够使得b,c数组成立。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200000+7;
#define LL long long
LL b[maxn],c[maxn],a[maxn];
int cnt[70];
int main()
{
int n;
LL sum = 0;
scanf("%d",&n);
for(int i = 1;i<=n;i++)scanf("%lld",&b[i]),sum+=b[i];
for(int i = 1;i<=n;i++)scanf("%lld",&c[i]),sum+=c[i];
if(sum%(2*n))
{
printf("-1\n");
return 0;
}
sum/=(2*n);
for(int i = 1;i<=n;i++)
{
if(c[i]+b[i]-sum<0 || (c[i]+b[i]-sum)%n)
{
printf("-1\n");
return 0;
}
a[i]=(c[i]+b[i]-sum)/n;
}
for(int i = 1;i<=n;i++)
for(int j = 0;j<60;j++)
if((a[i]>>j)&1)cnt[j]++;
for(int i = 1;i<=n;i++)
{
LL sumb = 0,sumc=0;
for(int j = 0;j<60;j++)
{
if((a[i]>>j)&1)
{
sumb+=cnt[j]*(1LL<<j);
sumc+=n*(1LL<<j);
}
else
sumc+=cnt[j]*1LL<<j;
}
if(sumb!=b[i] || sumc!=c[i])
{
printf("-1\n");
return 0;
}
}
for(int i = 1;i<=n;i++)
printf("%lld ",a[i]);
}
F. Anton and School
time limit per test
memory limit per test
input
output
Anton goes to school, his favorite lessons are arraystudying. He usually solves all the tasks pretty fast, but this time the teacher gave him a complicated one: given two arrays b and c of length n, find array a, such that:
where a and b means bitwise AND, while a or b
Usually Anton is good in arraystudying, but this problem is too hard, so Anton asks you to help.
Input
The first line of the input contains a single integers n (1 ≤ n ≤ 200 000) — the size of arrays b and c.
The second line contains n integers bi (0 ≤ bi ≤ 109) — elements of the array b.
Third line contains n integers ci (0 ≤ ci ≤ 109) — elements of the array c.
Output
If there is no solution, print - 1.
Otherwise, the only line of the output should contain n non-negative integers ai — elements of the array a. If there are multiple possible solutions, you may print any of them.
Examples
Input
4 6 8 4 4 16 22 10 10
Output
3 5 1 1
Input
5 8 25 14 7 16 19 6 9 4 25
Output
-1
