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 $B=[1,1,1,1]$
启发博客：http://www.cnblogs.com/stepping/p/7324970.html
官方题解：
1008 Rikka with Subset
签到题，大致的思想就是反过来的背包。
如果 B?i?? 是 B 数组中除了 B?0?? 以外第一个值不为 0 的位置，那么显然 i 就是 A 中的最小数。
现在需要求出删掉 i 后的 B 数组，过程大概是反向的背包，即从小到大让 Bj-=B?(j−i)??。
时间复杂度 O(nm)。
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<queue>
6 #include<cmath>
7 #include<string>
8 #include<map>
9 #include<vector>
10 using namespace std;
11
12 int b[10005];//最终结果b
13 int bb[10005];//目前得出的b
14 int a[10005];
15
16 int main()
17 {
18 int T,n,m,ans;
19 scanf("%d",&T);
20 while(T--)
21 {
22 scanf("%d%d",&n,&m);
23 for(int i=0;i<=m;i++)
24 scanf("%d",&b[i]);
25 memset(a,0,sizeof(a));
26 memset(bb,0,sizeof(bb));
27 bb[0]=1;
28 for(int i=1;i<=m;i++)
29 {
30 a[i]=b[i]-bb[i];
31 for(int j=1;j<=a[i];j++)//对bb进行更新
32 {
33 for(int k=m;k>=i;k--)//反着来避免已经加到结果里的数字再加一遍
34 bb[k]+=bb[k-i];
35 }
36 }
37 int flag=0;
38 for(int i=1;i<=m;i++)
39 for(int j=1;j<=a[i];j++)
40 {
41 if(flag++) printf(" ");
42 printf("%d",i);
43 }
44 printf("\n");
45 }
46 return 0;
47 }