hdu 5014__Number Sequence

Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2044    Accepted Submission(s): 798
Special Judge

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 = (a₀ ⊕ b₀) + (a₁ ⊕ b₁) +···+ (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 ≤ 10⁵), The second line contains a₀,a₁,a₂,...,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 b₀,b₁,b₂,...,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

 

Sample Output

20 1 0 2 3 4

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

題意：給出一個數字n，接下來輸入n+1個數字記爲數列a，這些數字不重複且大於等於從0到n。現要求另一數列b（特性與a數列一致）使得對應(a0+b0)^(a1+b1)^(a2+b2)^...^(an+bn)所得值最大。

想法：對於一個數a，從0到a的數字中與它異或得最大值的只有它的取反的數。例如5->101;從0到5裏應該取010，這樣就能夠利用貪心思想，從大到小，求出與之對應的那個值。

代碼如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
    int n,b[100000+10],a,ans[100000+10];
    while(~scanf("%d",&n)) {
        memset(ans,0,sizeof(ans));
        long long sum=0;
        for(int i=n;i>=0;--i) {
            if(ans[i]==0) {
                int t=i,k=0;
                while(t) {
                    b[k]=t%2;
                    k++;
                    t/=2;
                }
                for(int j=0;j<k;++j) {
                    b[j]=!b[j];
                    ans[i]+=b[j]*pow(2,j);
                }
                ans[ans[i]]=i;
                sum+=ans[i]^i;
            }
        }
        printf("%I64d\n",sum*2);
        for(int i=0;i<=n;i++) {
            scanf("%d",&a);
            if(i==0) {
                printf("%d",ans[a]);
            }
            else
                printf(" %d",ans[a]);
        }
        printf("\n");
    }
    return 0;
}