LS 2 Xor（数论）

Xor

For given multisets AandB, find minimum non-negative xwhichA⊕x=B.

Note that for A={a1,a2,…,an},A⊕x={a1⊕x,a2⊕x,…,an⊕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 nintegersa1,a2,…,an, which denote the set A.

The thrid line contains nintegersb1,b2,…,bn, which denote the set B.

(1≤n≤105,nis odd, 0≤ai,bi<230)

Output

The only integer denotes the minimum x. Print −1if no such xexists.

Sample input

30 1 31 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";  }}