题目链接
分析:这道题的关键点是要知道一个关于位运算的式子
a+b=(a|b)−(a & b)
这样的话,可以有以下的推导: ∑ni=1a[i]=sum ,那么 c[i]=∑Nj=1(a[i]+a[j])−b[i] ,然后推出 b[i]+c[i]=N∗a[i]+sum ,可以计算出 a[i]
还有一点需要注意,就是验证一下 a[i] 的正确性。这里也是要小技巧的。将整个式子拆成二进制的形式进行与和或的操作,这样能把长度为N的和式在 O(1) 的时间内求出来。
/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define offcin ios::sync_with_stdio(false)
#define sigma_size 26
#define lson l,m,v<<1
#define rson m+1,r,v<<1|1
#define slch v<<1
#define srch v<<1|1
#define sgetmid int m = (l+r)>>1
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x&-x)
#define bits(a) __builtin_popcount(a)
const int INF = ;
const ll INFF = ;
const double pi = acos(-);
const double inf = ;
const double eps = ;
const ll mod = +;
const int maxmat = ;
const ull BASE = ;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<)+(x<<)+c-'0';
}
/*****************************************************/
const int maxn = + ;
ll a[maxn], b[maxn], c[maxn];
int cnt[];
int main(int argc, char const *argv[]) {
int N; cin>>N;
ll sum = ;
for (int i = ; i <= N; i ++) cin>>b[i], sum += b[i];
for (int i = ; i <= N; i ++) cin>>c[i], sum += c[i];
if (sum % ( * N)) {puts("-1"); return ;}
sum = sum / N / ;
for (int i = ; i <= N; i ++) {
if ((b[i] + c[i] - sum) % N) {puts("-1"); return ;}
a[i] = (b[i] + c[i] - sum) / N;
}
for (int i = ; i <= N; i ++)
for (int j = ; j < ; j ++)
cnt[j] += a[i] >> j & ;
for (int i = ; i <= N; i ++) {
ll bb = , cc = ;
for (int j = ; j < ; j ++)
if (a[i] >> j & ) bb += cnt[j] * (l << j), cc += N * (l << j);
else cc += cnt[j] * (l << j);
if (bb != b[i] || cc != c[i]) {puts("-1"); return ;}
}
for (int i = ; i <= N; i ++) cout<<a[i]<<" ";
return ;
}