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

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)。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<vector>
using namespace std;

int b[10005];//最终结果b
int bb[10005];//目前得出的b
int a[10005];

int main()
{
    int T,n,m,ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<=m;i++)
            scanf("%d",&b[i]);
        memset(a,0,sizeof(a));
        memset(bb,0,sizeof(bb));
        bb[0]=1;
        for(int i=1;i<=m;i++)
        {
            a[i]=b[i]-bb[i];
            for(int j=1;j<=a[i];j++)//对bb进行更新
            {
                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
                    bb[k]+=bb[k-i];
            }
        }
        int flag=0;
        for(int i=1;i<=m;i++)
            for(int j=1;j<=a[i];j++)
            {
                if(flag++) printf(" ");
                printf("%d",i);
            }
        printf("
");
    }
    return 0;
}