Rikka with Subset（HDU 6092）

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 845    Accepted Submission(s): 412

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

22 31 1 1 13 31 3 3 1

 

Sample Output

1 21 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：b[i]代表的是a数组里一个或几个元素的和等于 i 的个数，现在告诉你a数组里有几个元素，给你b数组，要求a数组。比如，a里如果是1 1 2的话，b就是1 2 2 2 4。a里所有元素和为m。b里有m个元素，因为子集包括空集，所以b[0]恒等于1。

//思路：设b中除b[0]外第一个不为0的是b[i]，那么 i 必为a中元素，因为找不到比它更小的能构成它的数。

比如b[0]=1;b[1]=1;b[2]=1;b[3]=1; 那么根据b[1]=1我们可以推出a中一定有1。

然后，我们删去找出的这个元素，得到新的b数组。从小到大，b[j]=b[j]-b[j-i]; (j从i开始到m)

重复上述步骤，如果b1-bm都为0了，循环结束。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <queue>#include <vector>#include <algorithm>using namespace std;const int MAX = 1e4 + 100;int n, m;long long b[MAX];long long num[MAX];int main(){int i;int T;scanf("%d", &T);while (T--){memset(num, 0, sizeof(num));scanf("%d%d", &n, &m);for (int i = 0; i <= m; i++)scanf("%lld", &b[i]);int cnt = 0;while (true){int pos;for (i = 1; i <= m; i++){if (b[i] != 0){pos = i;num[cnt++] = i;break;}}if (i == m + 1)break;for (i = pos; i <= m; i++){b[i] = b[i] - b[i - pos];}}printf("%d", num[0]);for (i = 1; i < cnt; i++)printf(" %d", num[i]);printf("\n");}return 0;}