HDU5014：Number Sequence（贪心）

Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2442    Accepted Submission(s): 977
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

题意：找出符合题意的序列b，使得b0^a0 + b1^a1 + b2^a2 +... bn^an最大，其中bi<=n，输出最大值并输出b序列。

思路：异或最大，找出二进制互补的数字即可，要从大到小找。

写法一：

# include <stdio.h>
# include <string.h>
# define MAXN 100000
int a[MAXN+1], b[MAXN+1];
int fun(int num)
{
    int ans = 0;
    while(num)
    {
        num >>= 1;
        ++ans;
    }
    return ans;
}
int main()
{
    long long sum;
    int n;
    while(~scanf("%d",&n))
    {
        sum = 0;
        memset(b, -1, sizeof(b));
        for(int i=0; i<=n; ++i)
            scanf("%d",&a[i]);
        for(int i=n; i>=0; --i)
        {
            if(b[i] == -1)
            {
                int length = fun(i);
                int tmp = ((1<<length)-1)^i;
                b[i] = tmp;
                b[tmp] = i;
                sum += (i^tmp)<<1;
            }
        }
        printf("%lld\n",sum);
        for(int i=0; i<n; ++i)
            printf("%d ",b[a[i]]);
        printf("%d\n",b[a[n]]);

    }
    return 0;
}

写法二：

# include <stdio.h>
# include <string.h>
# define MAXN 100000
int a[MAXN+1], b[MAXN+1];
int main()
{
    long long n;
    while(~scanf("%lld",&n))
    {
        memset(b, -1, sizeof(b));
        for(int i=0; i<=n; ++i)
            scanf("%d",&a[i]);
        for(int i=n; i>=0; --i)
        {
            int ans=0, t=1, s=i;
            if(b[i] == -1)
            {
                while(s)
                {
                    ans += t*((s&1)^1);
                    t <<= 1;
                    s >>= 1;
                }
                b[i] = ans;
                b[ans] = i;
            }
        }
        printf("%lld\n",n*n+n);
        for(int i=0; i<n; ++i)
            printf("%d ",b[a[i]]);
        printf("%d\n",b[a[n]]);

    }
    return 0;
}