HDU 5014 number sequence (2014 ACM/ICPC Asia Regional Xi 'an online) question

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 5014

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

● AI in [0, N]
● AI = AJ (I = J)

For sequence a and sequence B, the integrating Degree t is defined as follows ("writable" denotes exclusive or ):

T = (A0 rjb0) + (A1 rjb1) + · + (an 1_bn)

(Sequence B shoshould also satisfy the rules described abve)

Now give you a number N and the sequence A. You shoshould calculate the maximum integrating Degree t and print the sequence B.
 
Inputthere 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,...,.
 
Outputfor 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

42 0 1 4 3

 
Sample output

201 0 2 3 4

 
Source2014 ACM/ICPC Asia Regional Xi 'an online

HDU has replaced _ int64 with long. The idea is very simple. Try to complete the values 0 and 1 of the two binary numbers, and give priority to a large number. However, you need to find the interval.

Code:

#include <iostream>#include <cstdio>using namespace std;__int64 n, a[100010];struct right{    __int64 s, r, l;}rt[1000];__int64 getNear(__int64 x){    __int64 z = 1;    while(x)    {        x >>= 1;        z <<= 1;    }    return z-1;}int main(){    while(~scanf("%I64d", &n))    {        __int64 m = n;        rt[0].r = m;        rt[0].s = getNear(m);        rt[0].l = rt[0].s-rt[0].r;        //cout << rt[0].l << " " << rt[0].r << " " << rt[0].s << endl;        __int64 cnt = 0;        while(1)        {            m = rt[cnt].l-1;            if(m  < 0) break;            cnt++;            rt[cnt].r = m;            rt[cnt].s = getNear(m);            rt[cnt].l = rt[cnt].s-rt[cnt].r;            //cout << rt[cnt].l << " " << rt[cnt].r << " " << rt[cnt].s << endl;        }        for(__int64 i = 0; i <= n; i++)            scanf("%I64d", &a[i]);            //a[i] = i;        __int64 t = 0;        for(__int64 i = 0; i <= n; i++)            for(__int64 j = 0; j <= cnt; j++)            {                if(a[i] >= rt[j].l && a[i] <= rt[j].r)                {                    //cout << rt[j].l << " " << rt[j].r << " " << rt[j].s << endl;                    //printf("%d ", rt[j].s-a[i]);                    a[i] = rt[j].s-a[i];                    t += rt[j].s;                    break;                }            }            printf("%I64d\n", t);            for(__int64 i = 0; i < n; i++)                printf("%I64d ", a[i]);            printf("%I64d\n", a[n]);    }    return 0;}

HDU 5014 number sequence (2014 ACM/ICPC Asia Regional Xi 'an online) question