HDU 5014 Number Sequence(贪心）

当时想到了贪心，但是不知为何举出了反列。。。。我是逗比，看了点击打开链接。才发现我是逗比。

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

42 0 1 4 3

 

Sample Output

201 0 2 3 4

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

枚举贪心即可。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cstdlib>#include<map>using namespace std;typedef long long LL;const int maxn=1e5+100;LL a[maxn];LL d[maxn];int main(){    LL n;    while(~scanf("%I64d",&n))    {        for(LL i=0;i<=n;i++)            scanf("%I64d",&a[i]);        memset(d,-1,sizeof(d));        LL ans=0;        for(LL i=n;i>=0;i--)        {            LL t=0;            if(d[i]==-1)            {                for(LL j=0;;j++)                {                    if(!(i&(1<<j)))  t+=(1<<j);                    if(t>=i)                    {                        t-=(1<<j);                        break;                    }                }                ans+=(i^t)*2;                d[i]=t;                d[t]=i;            }        }        printf("%I64d\n",ans);        for(LL i=0;i<=n;i++)        printf(i==n?"%I64d\n":"%I64d ",d[a[i]]);    }    return 0;}