Chip Factory

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1842 Accepted Submission(s): 833

Problem Description

John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces

Input

The first line of input contains an integer

Output

For each test case, please output an integer indicating the checksum number in a line.

Sample Input

2 3 1 2 3 3 100 200 300

Sample Output

6 400

给你n个整型数(n<=1e3)，现在可以从其中选出s1,s2,s3三个数字，要求求出(s1+s2)^s3的最大值，至多

1e3组测试数据。

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   #define MM(a,b) memset(a,b,sizeof(a));#define inf 0x3f3f3f3fusing namespace std;typedef long long ll;#define CT continue#define SC scanfconst int N=1e3+10;int ch[40*N][3];int a[N],sz,num[40*N],val[40*N];bitset<32> v;void add(int k){ int p=0; for(int i=30;i>=0;i--){ int id=v[i]; if(ch[p][id]==-1) ch[p][id]=++sz; p=ch[p][id]; num[p]+=k; } if(k!=-1) val[p]=v.to_ullong();//返回bitset的值}int query(){ int p=0; for(int i=30;i>=0;i--){ int id=v[i],ano=ch[p][!id]; if(ano!=-1&&num[ano]!=0) p=ano; else p=ch[p][id]; } return val[p];}int main(){ int cas,n; SC("%d",&cas); while(cas--){ MM(ch,-1);MM(num,0);sz=0; SC("%d",&n); for(int i=1;i<=n;i++){ SC("%d",&a[i]); v=a[i]; add(1); } int ans=0; for(int i=1;i<=n;i++){ v=a[i];add(-1); for(int j=i+1;j<=n;j++){ v=a[j];add(-1); v=a[i]+a[j]; ans=max(ans,(a[i]+a[j])^query()); v=a[j];add(1); } v=a[i];add(1); } printf("%d\n",ans); } return 0;}