Problem Description
There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:
● ai ∈ [0,n]
● ai ≠ aj( i ≠ j )
For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):
t = (a0 ⊕ b0) + (a1 ⊕ b1) +···+ (an ⊕ bn)
(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 ≤ 105), The second line contains a0,a1,a2,...,an.
Output
For each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b0,b1,b2,...,bn. There is exactly one space between bi
and bi+1(0 ≤ i ≤ n - 1). Don’t ouput any spaces after bn.
Sample Input
4 2 0 1 4 3
Sample Output
20 1 0 2 3 4 题意:求[0, n]的两个排列两两异或和的最大值 思路:算的上是一道规律题吧,拿5:101来说,能异或到的最大的数是111,那么就有(101,10)和(100,11)这两种,那么我们依次推下去求解 就是比如:0,1,2,3,4,5对应的值是:1,0,5,4,3,2,可以比较容易看到规律#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <vector> typedef __int64 ll; //typedef long long ll; using namespace std; const int maxn = 100005; ll ans[maxn]; int n; int cal(int m) { int len = 0; while (m) { m >>= 1; len++; } return len; } ll init(int m) { ll sum = 0; while (m >= 0) { int len = cal(m); ll Max = (ll)(1<<len) - 1; for (int i = Max-m; i <= m; i++) ans[i] = Max - i; sum += (ll)(m-(Max-m)+1) * Max; m = Max - m - 1; } return sum; } int main() { while (scanf("%d", &n) != EOF) { memset(ans, 0, sizeof(ans)); ll sum = init(n); printf("%I64d\n", sum); int x; for (int i = 0; i <= n; i++) { scanf("%d", &x); if (i == 0) printf("%I64d", ans[x]); else printf(" %I64d", ans[x]); } printf("\n"); } return 0; }