Number Sequence（二进制）-优快云博客

本文链接：https://blog.youkuaiyun.com/u014135990/article/details/88340244

Number Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1886 Accepted Submission(s): 561
Special 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 
  

题意：

给出 N ，后给出 N 个数，它是一个 0 ~ N 的排列，找出一个序列对应式子异或和最大。

思路：

将数变为二进制考虑，会发现每个数都会有对应有一个 “ 使之变为该二进制位数全为 1 ” 的数。这样子就能解决问题了，总和就是化为全为 1 时候对应的十进制值，对应的数就是这个和减去这个值的得数。

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

    return 0;
}