ACdream 1023 抑或

Xor

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

For given multisets Aand B, find minimum non-negative xx which A⊕x=BA⊕x=B

Note that for A={a1,a2,…,an}A={a1,a2,…,an} , A⊕x={a1⊕x,a2⊕x,…,an⊕x}. ⊕stands for exclusive-or.

Input

The first line contains a integer nn , which denotes the size of set AA (also for BB ).

The second line contains nn integers a1,a2,…,ana1,a2,…,an , which denote the set AA .

The thrid line contains nn integers b1,b2,…,bnb1,b2,…,bn , which denote the set BB .

(1≤n≤1051≤n≤105 , nn is odd, 0≤ai,bi<2300≤ai,bi<230 )

Output

The only integer denotes the minimum xx . Print −1−1 if no such xx exists.

Sample Input

3
0 1 3
1 2 3

Sample Output

2

Source

ftiasch

Manager

nanae

 

题意：集合A 与x抑或 得到集合B  输出最小的x  若不存在输出 -1

 

题解: 1.求抑或 满足交换律

        2.两个相同的数 抑或和为0

        3.x^0=x

        4.a^x=b 可以推出 a^b=x

       

 

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #include<cmath>
 #define ll long long
 #define pi acos(-1.0)
 #define mod 1000000007
 using namespace std;
 int ans1,ans2;
 int a[];
 int b[];
 int exm;
 int sum1=,sum2=;
 int main()
 {
     int n;
     while(scanf("%d",&n)!=EOF)
     {
         ans1=ans2=;
         sum1=;
         sum2=;
         for(int i=;i<=n;i++)
         {
          scanf("%d",&exm);
          a[i]=exm;
          ans1=ans1^exm;
         }
         for(int i=;i<=n;i++)
         {
          scanf("%d",&exm);
          b[i]=exm;
          sum2+=exm;
          ans2=ans2^exm;
         }
         ans1=ans1^ans2;
         for(int i=;i<=n;i++)
         {
          sum1=sum1+(ans1^a[i]);
         }
         if(sum1==sum2)
          cout<<ans1<<endl;
          else
          cout<<"-1"<<endl;
     }
     return ;
 }