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:

● ai ∈ [0,n]

● ai ≠ aj( i ≠ j )

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

t = (a0 ⊕ b0) + (a1 ⊕ b1) +···+ (an ⊕ bn)

(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,...,an.



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,...,bn. There is exactly one space between bi and bi+1(0
≤ i ≤ n - 1). Don’t ouput any spaces after bn.



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


题意：给出一个数字n，接下来输入n+1个数字记为数列a，这些数字不重复且大于等于从0到n。现要求另一数列b（特性与a数列一致）使得对应(a0+b0)^(a1+b1)^(a2+b2)^...^(an+bn)所得值最大。

想法：对于一个数a，从0到a的数字中与它异或得最大值的只有它的取反的数。例如5->101;从0到5里应该取010，这样就能够利用贪心思想，从大到小，求出与之对应的那个值。

代码如下：

#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;
}