HDU - 5014 Number Sequence

There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:

● a_i ∈ [0,n]
● a_i ≠ a_j( i ≠ j )

For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):

t = (a₀ ⊕ b₀) + (a₁ ⊕ b₁) +···+ (a_n ⊕ b_n)

(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.

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,...,a_n.

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,...,b_n. There is exactly one space between b_i and b_i+1(0 ≤ i ≤ n - 1). Don’t ouput any spaces after b_n.

4
2 0 1 4 3

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;
}