HDU5014 2014ACM-ICPC 亚洲区域赛西安赛区网络赛H题 Number Sequence

Number Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1885 Accepted Submission(s): 561

Special Judge

[align=left]Problem Description[/align]

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

● ai ∈ [0,n]
● ai ≠ aj( i ≠ j )

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

t = (a0 ⊕ b0) + (a1 ⊕ b1) +···+ (an ⊕ bn)

(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.

[align=left]Input[/align]

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,...,an.

[align=left]Output[/align]

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

[align=left]Sample Input[/align]

4
2 0 1 4 3

[align=left]Sample Output[/align]

20
1 0 2 3 4

这题在之前做时候完全想不到思路，直到比赛2小时脑子里突然闪现出一个YY思路，那就是如果对于一个0-n的序列，能构成的最大异或序列，需要这样构想，对于n，能构成的最大数字需要与n取反配对（n与n取反异或运算后所有二进制位置均为1，即最大），我们设n取反为m，那么n-1应当与m+1配对，依次类推。这样，我们得到了m-n所有的配对方式。然后，不断循环配对m-1与m-1取反区间内的数字，直到配对至0则完成全部配对，此时即为异或配对的最大值。。。输出即为答案。。。还是挺简单的嘛。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
using namespace std;
__int64 a[100005],pa[100005];
__int64 cal(__int64 pa)
{
__int64 ans=0,n=0;
while(pa>0)
{
a
=pa%2;
n++;
pa=pa/2;
}
__int64 tap=1;
for(int j=0;j<n;j++)
{
if(a[j]==0)
ans+=tap;
tap=tap*2;
}
return ans;
}
int main()
{
//   freopen("in.txt","r",stdin);
int n;
while(cin>>n)
{
for(int i=0;i<=n;i++)
scanf("%d",&pa[i]);
__int64 pat=cal(n),sum=0,psum=0;
sum+=(n-pat)*(pat+n+1);
for(int i=pat,j=n;i<=n;i++)
{
a[i]=j;
psum+=(i^j);
j--;
}
pat--;
while(pat>0)
{
__int64 item=cal(pat);
for(int i=item,j=pat;i<=pat;i++)
{
a[i]=j;
psum+=(i^j);
j--;
}
sum+=(pat-item)*(pat+item+1);
pat=item;
pat--;
}
cout<<psum<<endl;
for(int i=0;i<n;i++)
printf("%d ",a[pa[i]]);
printf("%d\n",a[pa
]);
}
return 0;
}