Number Sequence(バイナリ)

Number Sequence
Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1886    Accepted Submission(s): 561Special Judge
Problem Description
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 = (a
0 ⊕ b
0) + (a
1 ⊕ b
1) +···+ (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.
 
 
Input
There are multiple test cases. Please process till EOF.
For each case, the first line contains an integer n(1 ≤ n ≤ 10
5), The second line contains a
0,a
1,a
2,...,a
n.
 
 
Output
For each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b
0,b
1,b
2,...,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.
 
 
Sample Input
4
2 0 1 4 3
 
 
Sample Output
 
20
1 0 2 3 4
 
     タイトル:
     Nを与え,後にN個の数を与え,それは0~Nの配列であり,1つのシーケンスが式異和と最大に対応することを見出した.
 
     考え方:
     数をバイナリに変更すると、各数に「このバイナリビット数をすべて1にする」数が対応していることがわかります.このようにして問題を解決することができ,総和は全が1になったときに対応する十進法値となり,対応する数はこれとこの値を減算した得数となる.
 
     AC:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <cmath>

using namespace std;

typedef long long ll;

ll num[100005];
ll num1[100005];

ll Bit (ll ans) {
    ll bb = 0;
    while (ans) {
        ++bb;
        ans /= 2;
    }
    return bb;
}

int main() {

    int n;

    while (~scanf("%d", &n)) {
        for (int i = 0; i <= n; ++i) {
            scanf("%I64d", &num[i]);
        }

        memset(num1, -1, sizeof(num1));

        ll sum = 0;
        for (ll i = n; i >= 0; --i) {
            if (num1[i] == -1) {
                ll bb = Bit(i);
                ll ans = pow(2, bb) - 1;
                sum += ans * 2;
                num1[i] = ans - i;
                num1[ans - i] = i;
            }
        }

        if (num1[0] == -1) num1[0] = 0;

        printf("%I64d
", sum);
        for (int i = 0; i <= n; ++i) {
            printf("%I64d", num1[num[i]]);
            i == n ? printf("
") : printf(" ");
        }
    }

    return 0;
}