2017杭电多校第五场Rikka with Subset

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1440    Accepted Submission(s): 721

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 $B=[1,1,1,1]$

 

Source

2017 Multi-University Training Contest - Team 5

 

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思路：动态规划+思维 因为已知了集合B要求集合A的序列，显然空集与全集的数量都为1，所以B0和Bm都为1 集合A中1的数量就等于B1，那么B2便可以由B1推出（排列组合的思想），B3可有B2推出，以此类推，采用01背包为题解决

#include <iostream>
#include<algorithm>
#include<string.h>
#include<stdint.h>
using namespace std;
const int maxn=10005;

int a[maxn],b[maxn],c[maxn],dp[maxn];
//dp[i]表示：加和为i的子集个数

int main()
{
    int t;
    scanf("%d",&t);
    int n,m;
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        memset(dp,0,sizeof(dp));

        dp[0]=1;
        for(int i=0;i<=m;i++)
        {
          scanf("%d",&b[i]);
        }
        int p=0,sum=0;
        for(int i=1;i<=m;i++)
        {
            c[i]=b[i]-dp[i];//A序列中值为i的个数
            for(int j=0;j<c[i];j++)
            {
                a[p++]=i;//对A序列赋值
                for(int k=m;k>=i;k--)
                {//处理成01背包问题
                    dp[k]+=dp[k-i];//和为k的子集个数相加去更新B序列

                }
            }

        }
        for(int i=0;i<p-1;i++)
        {
            printf("%d ",a[i]);

        }
        printf("%d\n",a[p-1]);
    }
    return 0;
}