HDU 5014 Number Sequence(贪心） - c++编程基础

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 = (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

Sample Output

20
1 0 2 3 4

Source 2014 ACM/ICPC Asia Regional Xi'an Online
枚举贪心即可。

#include
  
   
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        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<
        
         =i) { t-=(1<