Codeforces Round #177 (Div. 2)---E. Polo the Penguin and XOR operation (greedy)

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

For permutation p?=?p0,?p1,?...,? pn, Polo have 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

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 is several suitable permutations, you is allowed to print any of them.
Sample Test (s)
Input

4

Output

20
0 2 1) 4 3

Observations can be found, two-digit XOR or later, if the binary each bit is 1, then it must be the largest, so our strategy is to enumerate each 2 i ? 1 , of course, starting from the one that is greater than or equal to N, and then each time to find the satisfaction that meets A+B= 2 i ? 1 The number

/************************************************************************* > File name:cf-177-e.cpp > Aut Hor:alex > Mail: [email protected] > Created time:2015 April 08 Wednesday 16:01 33 seconds ******************************* *****************************************/#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 DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;intper[1001000];intMain () {intN while(~scanf("%d", &n)) {intUse1 =0, Use2 =0; for(inti =1; I <= N;        ++i) {use1 ^= i; } LL ans =0;intm = n;intHigh =0; while(High < N) {high =2* High +1; }intUse 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(inti =1; I <= m; ++i) {printf("%d", Per[i]); }printf("\ n"); }return 0;}

Codeforces Round #177 (Div. 2)---E. Polo the Penguin and XOR operation (greedy)