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2i ?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=2i ?1的数
/*************************************************************************
> File Name: CF-177-E.cpp
> Author: ALex
> Mail: [email protected]
> 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;
}