ACdream 1023 抑或

Xor

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

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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

Sample Output

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 ;

 }

巴特西