贪心 HDU 5014





Problem Description

There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:

● a _i ∈ [0,n] 
● a _i ≠ a _j( i ≠ j )

For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):

t = (a ₀ ⊕ b ₀) + (a ₁ ⊕ b ₁) +···+ (a _n ⊕ b _n)

(sequence B should also satisfy the rules described above)

Now give you a number n and the sequence a. You should calculate the maximum integrating degree t and print the sequence b.

 

Input

There are multiple test cases. Please process till EOF. 

For each case, the first line contains an integer n(1 ≤ n ≤ 10 ⁵), The second line contains a ₀,a ₁,a ₂,...,a _n.

 

Output

For each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b ₀,b ₁,b ₂,...,b _n. There is exactly one space between b _i and b _i+1 (0 ≤ i ≤ n - 1). Don’t ouput any spaces after b _n.

 

Sample Input

   
   
   
   

    
    
    
    4
   
   
   
   
   
   
   
   

    
    
    
    2 0 1 4 3
   
   
   
   
   
   
   
   

    
    
    
    思路：贪心:要明白二进制异或的贪心规则，当2个数异或时最高位是1与0，可以算出最大数。（从最大的一个数开始找能配对使他们的异或值最大的一个数即可）
    
    
    
    
#include <stdio.h>
#include <string.h>

int str[1000005],judge[1000005];
int main()
{
    int n,i,itemp,j,sum;
    //freopen("e:\\in.txt","r",stdin);
    while(scanf("%d",&n)==1)
    {   long long ans=0;
        memset(judge,-1,sizeof(judge));   //判断数组

        for(itemp=0;itemp<=n;itemp++)
            scanf("%d",&str[itemp]);//输入数组

         for(i=n;i>=0;i--)
         {

             if(judge[i]==-1)
             {   sum=0;
                 for(itemp=0;;itemp++)
                 {
                     if(!(i&(1<<itemp)))   sum+=1<<itemp;

                     if(sum>=i)   break;
                 }
                 sum-=(1<<itemp);
                 ans+=(i^sum)*2;

                 judge[i]=sum;
                 judge[sum]=i;

             }

         }

         printf("%lld\n",ans);
         for(j=0;j<n;j++)
            printf("%d ",judge[str[j]]);
           printf("%d\n",judge[str[n]]);
    }
    return 0;
}
关键：
    
    
    
    

      1.2个数组之间的嵌套。把数组str嵌套在judge,直接把所对应的计算出来即可。 
    
    
    
    
    

      2.对2进制异或内容与数字之间规律的掌握。 
    
    
    
    
    

      3.判断数组judge的判断内容，if（judge[i]）与if(judge[i]==-1)这是导致初写程序错误的主要原因。