HDU - 6092 Rikka with Subset

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−AnA1−An and their sum is mm. Then for each subset SS of AA, Yuta calculates the sum of SS. 

Now, Yuta has got 2n2n numbers between [0,m][0,m]. For each i∈[0,m]i∈[0,m], he counts the number of iis he got as BiBi. 

Yuta shows Rikka the array BiBi and he wants Rikka to restore A1−AnA1−An. 

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

 

Input

The first line contains a number t(1≤t≤70)t(1≤t≤

Sample Input

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

70), the number of the testcases. 

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

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

 

Output

For each testcase, print a single line with nn numbers A1−AnA1−An. 

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

 

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：一个集合A中有n个元素，集合A中所有的元素和为m,求出集合A所有子集的和,集合B中元素b[i]表示集合A的子集中所有和为i 的个数，给出集合B求出集合A

解题思路：逆向dp,我们设dp[i]表示集合A的子集中和为i的个数，则可以这样求出dp[i]

	for(int i=1;i<=n;i++)
	{
		for(int j=m;j>=a[i];j--)
			dp[j]+=dp[j-a[i]];
	}

 根据求dp[i]的方法可以逆向dp来求a[i]，我们需要找出集合A的第一个元素从1开始从前往后找，那么第一个b[i]不等于0的i一定是集合A中的最小的那个元素，因为b[i]是集合A的子集中和为i的个数，集合A中最小的那个元素单独一个子集和最小，这时就很容易找到所有集合A的元素，然后逆向dp,每次都把最小的元素加进去，这样最后输出的就是字典序最小的集合A

AC代码

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1e4+10;
typedef long long LL;
LL dp[maxn],b[maxn];
int a[55];
int main()
{
	int t,n,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=0;i<=m;i++)
			scanf("%lld",&b[i]);
		for(int i=1;i<=m;i++)
		{
			if(b[i])
			{
				a[1]=i;//找集合A最小的那个元素 
				break;
			}
		}
		int cnt=2;
		for(int i=1;i<=n;i++)////从第一个数反过来DP
		{
			for(int j=a[i],flag=0;j<=m;j++)
			{
				b[j]-=b[j-a[i]];//去掉A的第i和元素时和为j的个数 
				if(b[j]>0&&flag==0)
				{
					a[cnt++]=j;///每次都要把最小的j且b[j]!=0的数加入a中
					flag=1;
				}
			}
		}
		for(int i=1;i<=n;i++)
		{
			if(i==n)
				printf("%d\n",a[i]);
			else
				printf("%d ",a[i]);
		}
	}
	return 0;
}