Xor
For given multisets A and B , find minimum non-negative x which A⊕x=B .
Note that for A={a 1 ,a 2 ,…,a n } , A⊕x={a 1 ⊕x,a 2 ⊕x,…,a n ⊕x} . ⊕ stands for exclusive-or.
Input
The first line contains a integer n , which denotes the size of set A (also for B ).
The second line contains n integers a 1 ,a 2 ,…,a n , which denote the set A .
The thrid line contains n integers b 1 ,b 2 ,…,b n , which denote the set B .
(1≤n≤10 5 , n is odd, 0≤a i ,b i <2 30 )
Output
The only integer denotes the minimum x . Print −1 if no such x exists.
Sample input
3
0 1 3
1 2 3
Sample output
2
Source
Contest #3 on acmdream.net by ftiasch
思路:如果A^x=B 那么统计A,B集合中所有数各个位的1的个数和作比较,如果不等那么x的该位必须为1,因为0异或操作不改变任何结果
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int mm=100210;
const int bit=31;
int n;
bool p[mm];
int a[mm],b[mm],ca[bit],cb[bit];
void calculate(int *c,int *cc)
{
for(int i=0;i<n;i++)
for(int j=0;j<bit;j++)
cc[j]+=(c[i]>>j)&1;
}
int main()
{
while(cin>>n)
{ memset(ca,0,sizeof(ca));
memset(cb,0,sizeof(cb));
for(int i=0;i<n;i++)
cin>>a[i];
for(int i=0;i<n;i++)
cin>>b[i];
calculate(a,ca);calculate(b,cb);
int x=0;
for(int i=0;i<bit;i++)
if(ca[i]^cb[i])x|=1<<i;
bool yes=1;
for(int i=0;i<n;i++)
a[i]^=x;
sort(a,a+n);sort(b,b+n);
for(int i=0;i<n;i++)
if(a[i]^b[i])yes=0;
cout<<(yes?x:-1)<<"\n";
}
}