ACdream 1023 抑或

Posted 2020-07-15 半根毛线code

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

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

       

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<queue>
 5 #include<stack>
 6 #include<cmath>
 7 #define ll long long 
 8 #define pi acos(-1.0)
 9 #define mod 1000000007
10 using namespace std;
11 int ans1,ans2;
12 int a[100005];
13 int b[100005];
14 int exm;
15 int sum1=0,sum2=0;
16 int main()
17 {
18     int n;
19     while(scanf("%d",&n)!=EOF)
20     {
21         ans1=ans2=0;
22         sum1=0;
23         sum2=0;
24         for(int i=1;i<=n;i++)
25         {
26          scanf("%d",&exm);
27          a[i]=exm;
28          ans1=ans1^exm;
29         }
30         for(int i=1;i<=n;i++)
31         {
32          scanf("%d",&exm);
33          b[i]=exm;
34          sum2+=exm;
35          ans2=ans2^exm;
36         }
37         ans1=ans1^ans2;
38         for(int i=1;i<=n;i++)
39         {
40          sum1=sum1+(ans1^a[i]);
41         }
42         if(sum1==sum2)
43          cout<<ans1<<endl;
44          else
45          cout<<"-1"<<endl;     
46     }
47     return 0;
48 }