Codeforces Round #177 (Div. 二)-E. Polo the Penguin and XOR operation（贪心）

Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.

For permutation p = p0, p1, …, pn, Polo has defined its beauty — number .

Expression means applying the operation of bitwise excluding “OR” to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as “^” and in Pascal — as “xor”.

Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.
Input

The single line contains a positive integer n (1 ≤ n ≤ 106).
Output

In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.

If there are several suitable permutations, you are allowed to print any of them.
Sample test(s)
Input

4

Output

20
0 2 1 4 3

观察可以发现，两个数异或以后，如果二进制每一位都为1，那么一定最大，所以我们的策略是，枚举每一个$2^{i}$$- 1$,当然是从大于等于n的那个开始，然后每次去找符合的满足 $A + B = 2^{i}$$- 1$的数

/*************************************************************************
    > File Name: CF-177-E.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年04月08日 星期三 16时01分33秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int per[1001000];

int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        int use1 = 0, use2 = 0;
        for (int i = 1; i <= n; ++i)
        {
            use1 ^= i;
        }
        LL ans = 0;
        int m = n;
        int high = 0;
        while (high < n)
        {
            high = 2 * high + 1; 
        }
        int use;
        while (high)
        {
            use = n;
            while (use > 0 && high - use <= n)
            {
                ans += high;
                per[use] = high - use;
                use2 ^= (high - use);
                --use;
            }
            high >>= 1;
            n = use;
        }
        per[0] = (use2 ^ use1);
        ans += per[0];
        printf("%I64d\n", ans);
        printf("%d", per[0]);
        for (int i = 1; i <= m; ++i)
        {
            printf(" %d", per[i]);
        }
        printf("\n");
    }
    return 0;
}