Rikka with Subset（hdu6092）

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has n positive A1−An and their sum is m. Then for each subset S of A, Yuta calculates the sum of S.

Now, Yuta has got 2n numbers between [0,m]. For each i∈[0,m], he counts the number of is he got as Bi.

Yuta shows Rikka the array Bi and he wants Rikka to restore A1−An.

It is too difficult for Rikka. Can you help her?  

 

Input

The first line contains a number t(1≤t≤70), the number of the testcases.

For each testcase, the first line contains two numbers n,m(1≤n≤50,1≤m≤104).

The second line contains m+1 numbers B0−Bm(0≤Bi≤2n).

 

Output

For each testcase, print a single line with n numbers A1−An.

It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.

 

Sample Input

2
2 3
1 1 1 1
3 3
1 3 3 1

 

Sample Output

1 2
1 1 1
Hint

In the first sample, $A$ is $[1,2]$. $A$ has four subsets $[],[1],[2],[1,2]$ and the sums of each subset are $0,1,2,3$. So $B=[1,1,1,1]$

题意：一共有T组测试样例，然后n和m，有m+1个数代表B数列，表示为A数列子集的和为i的有多少种；
解题思路：很多个较小的数字随机组合会求出多个很大的数字，所以从B0向Bm推导，在每求出A序列的一部分这个过程中，更新后续的B序列，更新完的B[i]就是 i 在A序列中出现的次数。 
分析完后，主要的难点就是怎么去让已求出来的A序列随机组合，更新后续的B序列直接减就可以了。看成01背包问题，让m为背包去装 i，初始值为dp[0] = 1，由于i依次增大，A子集随机组合不会重复

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
int n,m;
int ans[10005],tmp[10005],dp[10005];
int da[10005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {


        scanf("%d %d",&n,&m);
        for(int i=0;i<=m;i++)
        scanf("%d",&da[i]);
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));
        memset(tmp,0,sizeof(tmp));
        dp[0]=1;
        int p=0;
        for(int i=1;i<=m;i++)
        {
            tmp[i]=da[i]-dp[i];
            for(int j=0;j<tmp[i];j++)
            {
                ans[p]=i;
                p++;
                for(int k=m;k>=i;k--)
                {
                    dp[k]+=dp[k-i];//和为k的子集不断更新
                }
            }
        }
        for(int i=0;i<n-1;i++)
        printf("%d ",ans[i]);
        printf("%d\n",ans[n-1]);
    }
    return 0;
}