hdu 5244 inverse (上海大都会赛) （分治算法）

最新推荐文章于 2019-04-18 22:43:57 发布

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#分治 #上海大都会赛

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本文介绍了一种基于位运算和特定函数f的加密方法，并通过分治策略实现逆向解析，以恢复原始数组。该方法利用了特殊的数学性质，使得逆向过程变得可行。

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nverse

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 110 Accepted Submission(s): 44

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.

b i = \sum 0 \leq j < n f ((i or j) xor i) a j

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

Input

The first line contains an integer

T≤5), denoting the number of the test cases.

For each test case, the first line contains an integer

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

Sample Input

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;
}