看着样例推算了一下...得出的一个方法..让每个数与其选择的数XOR后为全1并且1尽可能的多.....
如样例
0 1 2 3 4
XOR 0 2 1 4 3
-----------------------
0 3 3 7 7
Program:
#include<iostream>
#include<stack>
#include<queue>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<cmath>
#define ll long long
#define oo 1000000007
#define MAXN 1000005
using namespace std;
int n;
ll ans[MAXN];
int main()
{
int i;
ll x,sum;
while (~scanf("%d",&n))
{
x=1;
while (x<=n) x<<=1;
x--;
memset(ans,-1,sizeof(ans));
for (i=n;i>=0;i--)
{
if (ans[i]!=-1) continue;
while ((x^i)>n || ans[x^i]!=-1) x>>=1;
ans[x^i]=i,ans[i]=x^i;
}
sum=0;
for (i=0;i<=n;i++) sum+=i^ans[i];
printf("%I64d\n",sum);
for (i=0;i<=n;i++) printf("%I64d ",ans[i]);
printf("\n");
}
return 0;
}