HDU 6092 17多校5 Rikka with Subset（dp+思维）

Posted 2020-10-02 Annetree的博客

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 }