題意:給出兩個陣列b,c,構造a陣列滿足條件.

首先需要知道一個推論:aANDb+aORb=a+b,這個按位比較很顯然是相等的. 然後把原式的bi和ci加起來就得到了:
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪b.1+c1=na1+∑ab2+c2=na2+∑a⋮bn+cn=nan+∑a
這麼一個n元1次方程解出來也很容易,然後就是帶入按位check, 按照ai在當前位的奇偶性討論即可.

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
 

#include <algorithm>
#include <stack>
#define Clear(x,y) memset (x,y,sizeof(x))
#define Close() ios::sync_with_stdio(0)
#define Open() freopen ("more.in", "r", stdin)
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<long long, int>
#define pb push_back
 

#define mod 1000000007
template <class T>
inline bool scan (T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return 0; //EOF
    while (c != '-' && (c < '0' || c > '9') ) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0'
 
 && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}
using namespace std;
#define maxn 200005

long long b[maxn], c[maxn], f[maxn], a[maxn], ans[maxn];
int n;

int main () {
    //Open ();
    Close ();
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> b[i];
    long long sum = 0;
    for (int i = 1; i <= n; i++) {
        cin >> c[i];
        f[i] = b[i]+c[i];
        sum += f[i];
    }
    a[1] = (f[1]*2*n-sum);
    if (a[1]%(1LL*2*n*n) != 0) {
        cout << "-1" << endl;
        return 0;
    }
    a[1] /= (1LL*2*n*n); ans[1] = a[1];
    if (a[1] < 0) {
        cout << -1 << endl;
        return 0;
    }
    for (int i = 2; i <= n; i++) {
        if ((f[i]-f[1])%n != 0) {
            cout << -1 << endl;
            return 0;
        }
        a[i] = a[1]+(f[i]-f[1])/n;
        if (a[i] < 0) {
            cout << -1 << endl;
            return 0;
        }
        ans[i] = a[i];
    }
    for (int bit = 1; bit <= 33; bit++) {
        int num = 0;
        for (int i = 1; i <= n; i++) {
            num += (a[i]&1);
        }
        for (int i = 1; i <= n; i++) {
            if (a[i]&1) b[i] -= num, c[i] -= n;
            else c[i] -= num;
            if ((b[i]&1) || (c[i]&1)) {
                cout << -1 << endl;
                return 0;
            }
            b[i] >>= 1;
            c[i] >>= 1;
        }
        for (int i = 1; i <= n; i++) a[i] >>= 1;
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << " "; cout <<endl;
    return 0;
}