Spring-1-H Number Sequence（HDU 5014）解题报告及测试数据

Number Sequence

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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

题解：

1. 要使xor值最大，那么应该是xor的结果每一位上全部为1，例如n=3,00与11,01与10,10与01,11与00， 贪心算法，从最大的那个数匹配开始，尽量满足大的数字。 

2. i从n开始， 所以1不断左移，直到匹配的数比不比i小为止。判断某一位上是0还是1，需要用到&运算符，若i的第k位上是0，那么i&(1<<k)的值就是0。     

以下是代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <sstream>
#include <queue>
#include <stack>
#include <stack>
#include <map>
using namespace std;
 
#define F(i,s,e) for(int i = s;i<e;i++)
#define FA(i,s,e) for(int i=s;i>e;i--)
#define ss(x) scanf("%d",&x)
#define s64(x) scanf("%I64d",&x)
#define write() freopen("1.in","r",stdin)
#define W(x) while(x)
 
typedef long long LL;
LL a[100005];
LL b[100005];
 
int main(){
    write();
    LL sum,n;
    W(s64(n)!=EOF){
        sum=0;
        for(LL i=0;i<=n;i++)s64(a[i]);
        for(LL i=0;i<=n;i++)b[i]=-1;
        for(LL i=n;i>=0;i--){
            if(b[i]!=-1)continue;
            LL t=0,k=0;
            W(1){
                if((i&(1<<k))==0)t+=1<<k;//依次按位取反
                if(t>=i){         //直到比i大
                    t-=1<<k;
                    break;
                }
                k++;   
            }
            b[i]=t;
            b[t]=i;
            sum+=(t^i)*2;
        }
        printf("%I64d
",sum);
        for(LL i=0;i<n;i++)
            printf("%I64d ",b[a[i]]);//输出对应位置的值
        printf("%I64d
",b[a[n]]);
    }
}