Rikka with Subset HDU

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 nn 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

i s 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≤10e4)

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]

题目大意：有两个数组 A[] 和 B[] ,给出两个数 n，m，A[] 数组中有 n  个数，所有的数之和为 m 。接下来一行输入B数组，B[x] 数组中装的是 A[] 中和为 x 的自己的个数.  样例：A[] 数组为 [1,2], 那么它的子集为 [] ,[1], [2], [1,2].这些子集的和为 0，1，2，3。 所以 B[0]=1 ,  B[1]=1, B[2]=1, B[3]=1;现在给出 B 数组，求 A 数组。

   第二个样例的意思是： 0 1 2 3 的子集个数分别有1 3 3 1个,原集合a一共有3个元素

题目分析：  

                       cnt[i]表示子集的和为 i 的子集个数

               sum[i]表示求到当前位置之前子集和为i的子集个数.

      首先我们发现原集合中0的个数是好求的,      2^num[0]=cnt[0]

          

num[i]*sum[0]+sum[i]=cnt[i],也就是说一共和为 i 的子集个数等于集合中数字 i 的个数乘和为0的子集个数再加上之前那些由>0&&<i的元素构成的子集中和为i的个数

然后我们每次求出一个num[i]后,用完全背包的思想将这num[i]个i一个一个的加入到集合中,用完全背包的思想更新sum数组就行了。

代码如下：

#include <bits/stdc++.h>using namespace std;const int maxn = 5e5+500;int n,m;long long cnt[maxn];long long sum[maxn];int num[maxn];///A集合中值为 i 的数量int main(){    int t;    scanf("%d",&t);    while (t--)    {        memset(sum,0,sizeof(sum));        memset(num,0,sizeof(num));        scanf("%d%d",&n,&m);        for (int i=0; i<=m; ++i)            scanf("%lld",&cnt[i]);        num[0]=0;        sum[0]=cnt[0];        while((1<<num[0])<cnt[0])///2^num[0]<cnt[0]            num[0]++;        for(int i=1; i<=m; ++i)        {            num[i]=(cnt[i]-sum[i])/sum[0];///num[i]*sum[0]+sum[i]=cnt[i]            for (int j=1; j<=num[i];++j) ///一个一个的加入几个            {                for (int k=m; k>=i;--k) ///完全背包思想更新sum                {                    sum[k]+=sum[k-i];                }            }        }        vector<int> vec;        for (int i=0;i<=m;++i)        {            for (int j=0; j<num[i]; ++j)                vec.push_back(i);        }        for (int i=0; i<vec.size(); ++i)            printf("%d%c",vec[i],i==(vec.size()-1)?'\n':' ');    }    return 0;}