hdu 5014 Number Sequence 找规律 | 贪心 2014 ACM/ICPC Asia Regional Xi'an Online

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5014

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 = (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.

 

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,...,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 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.

 

Sample Input

42 0 1 4 3

 

Sample Output

201 0 2 3 4

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

线下 AC的，比赛的时候其实已经找到规律了，然后自己二呵呵的把binary数组初值赋错了，一直没发现....粗糙党

不过后来看正解是贪心...但我们的规律也是按贪心找的...

思路：

     如果两个数的二进制位数不同，这两个数的异或值就会越大，所以尽量使   a^b == 2^x - 1，这样我们得到了一个基本关系。

  

a:   0  1  2  3  4   5  6  7  8  9  10 11  12  13  14  15  16 17 ...
b:   0  2  1  4  3  10  9  8  7  6  5  20  19  18  17  16  15 14 ...

可以看出从17~11是递增，10~5也是递增的，4~3是递增的，2~1递增，如果a是上限，那么直接找2^x - 1 >= a，这里=a ，只有第一次累加的时候，然后一次往回加，加到小于上一次的最小值....重复这个过程...我用了low和high来完成累加的过程，并标记每一次的上下限

代码：

#include <algorithm>#include <cstdlib>#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <cctype>#include <cmath>#include <stack>#include <queue>#include <list>#include <map>#include <set>using namespace std;#define min2(x, y)     min(x, y)#define max2(x, y)     max(x, y)#define min3(x, y, z)  min(x, min(y, z))#define max3(x, y, z)  max3(x, max(y, z))#define clr(x, y)      memset(x, y, sizeof(x))#define fr(i,n)        for(int i = 0; i < n; i++)#define fr1(i,n)       for(int i = 1; i < n; i++)#define upfr(i,j,n)    for(int i = j; i <= n; i++)#define dowfr(i,j,n)   for(int i = n; i >= j; i--)#define scf(n)         scanf("%d", &n)#define ptf(n)         printf("%d",n)#define ptfl(n)        printf("%I64d",n)#define ptfs(s)        printf("%s",s)#define ptk()          printf(" ")#define ptln()         printf("\n")#define srt(a,n)       sort(a,n)#define LL long long#define pi acos(-1.0)#define inf 1 << 31-1#define eps 0.00001#define maxn 100005int arr[maxn];int ans[maxn];int binary[maxn];int vis[maxn];int main(){    //freopen("in.txt","r",stdin);    binary[0] = 1;    //2的0次是1，开始自己写成2了，2333    fr1(i,20)    binary[i] = binary[i-1] * 2;    int n;    while(scf(n) !=EOF )    {        clr(vis,0);        clr(ans,0);        clr(arr,0);        fr(i,n+1)        scf(arr[i]);        int tmp;        fr(i,20)        {            if(binary[i] > n)                {                    tmp = binary[i];                    break;                }        }        //ptf(tmp);        int low,high;        low = tmp - 1 - n;        ans[n] = low;        high = n;        dowfr(i,0,n-1)        {            if(ans[i+1]+1 == n)                ans[i] = n;            else if(ans[i+1] + 1 < high)                ans[i] = ans[i+1] + 1;            else            {                fr(j,20)                {                    if(binary[j] > i)                    {                        high = low ;                        low = binary[j] - 1 - i;                        break;                    }                }                ans[i] = low;            }        }        __int64 sum = 0;        fr(i,n+1)        sum += i ^ ans[i];        ptfl(sum);        ptln();        ptf(ans[arr[0]]);        upfr(i,1,n)           ptk(),ptf(ans[arr[i]]);        ptln();    }    return 0;}