nverse

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Mike has got a huge array b,
and he is told that the array is encrypted.

The array is encrypted as follows.

Let ai(0≤i<n) be
the i-th
number of this original array.

Let bi(0≤i<n) be
the i-th
number of this encrypted array.

Let n be
a power of 2,
which means n=2k.

The bi is
calculated as following.

bi=∑0≤j<nf((i or j) xor i)aj

f(x) means,
if the number of 1 in
the binary of x is
even, it will return 1,
otherwise 0.

Mike want to inverse the procedure of encryption.

Please help him recover the array a with
the array b.

Input

The first line contains an integer T(T≤5),
denoting the number of the test cases.

For each test case, the first line contains an integer k(0≤k≤20),

The next line contains n=2k integers,
which are bi respectively.

It is guaranteed that, ai is
an integer and 0≤ai≤109.

Output

For each test case, output ‘‘Case #t:‘‘ to represent this is the t-th case. And then output the array a.

Sample Input

2
0
233
2
5 3 4 10

Sample Output

Case #1: 233
Case #2: 1 2 3 4

分治。吧矩阵构造出来们发现有规律 ，然后就可以用分治求解了。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int mmax = (1<<21);
const int inf = 0x3fffffff;
typedef __int64 LL;
LL b[mmax],a[mmax];
void cdq(int l,int r)
{
    if(l==r)
    {
        a[l]=b[l];
        return ;
    }
    int mid=(l+r)>>1;
    for(int i=l;i<=mid;i++)
    {
        LL tmp=b[i]+b[i+(r-l+1)/2];
        b[i]=(tmp-(b[r]-b[mid]))/2;
        b[i+(r-l+1)/2]-=b[i];
    }
    cdq(l,mid);
    cdq(mid+1,r);
}
int get(int x)
{
    int cnt=0;
    while(x)
    {
        if(x&1)
            cnt++;
        x/=2;
    }
    return cnt;
}
int c[mmax];
void test(int k)
{
    for(int i=0;i<(1<<k);i++)
    {
        int sum=0;
        for(int j=0;j<(1<<k);j++)
            if( get((i|j)^i ) %2==0 )
                sum+=a[j+1];
        b[i+1]=sum;
    }
}
int main()
{
    int t,k,ca=0;
    cin>>t;
    while(t--)
    {
        scanf("%d",&k);
        for(int i=1;i<=(1<<k);i++)
            scanf("%I64d",&b[i]);
//        for(int i=1;i<=(1<<k);i++)
//            scanf("%d",&a[i]);
//        test(k);
        cdq(1,(1<<k));
        printf("Case #%d: ",++ca);
        for(int i=1;i<=(1<<k);i++)
            printf("%I64d%c",a[i],i==(1<<k)?'\n':' ');
        //test(k);
    }
    return 0;
}

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