hdu 6092 Rikka with Subset（逆向01背包+思维）

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1122    Accepted Submission(s): 541

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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题目大意：

有一个数列 a[] ,长度（n<=50）。b[i] 表示元素和为 i 的集合个数。给你一个数列 b[] ，长度（m<=10000），让你求 a[]，并按照其字典序最小输出。

题解：

网上有说是01背包，然而我也不知道自己写了点什么，大概是01背包吧

正常的背包：告诉你这个重量的物品个数，让你装包

逆向背包：这题告诉你，得到的这个重量有多少种，求各个重量的物品个数

 

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 t,k;
int  n,m;
int bb[10005],b[10005],a[10005];
int main()
{
    scanf("%d",&t);

    while(t--)
    {
       scanf("%d%d",&n,&m);
       memset(a,0,sizeof(a));//a[i]表示第i个数字的个数
       memset(bb,0,sizeof(bb));//bb[k]表示当1～k-1中数字个数确定后，凑到和为k的种数，不够就表示，需要单独k这个数字来凑

       for(int i=0;i<=m;i++)
           scanf("%d",&b[i]);

       k=1; bb[0]=1;
       while(k<=m)
       {
           a[k]=b[k]-bb[k]; //这就是因为1～k-1个数确定后，能凑到和为k的种数，不够的说明a序列中有b[k]-bb[k]个数的k

            for(int j=1;j<=a[k];j++)
            {
                for(int i=m;i>=k;i--) //反着来，避免已经加到结果里的数字再加一遍，这里有01背包的感觉
                   bb[i]+=bb[i-k];
            }
           k++;
       }

       int tot=0;  //输出
       for(int i=1;i<=m;i++)
          for(int j=1;j<=a[i];j++)
           {
               if (tot++) printf(" ");
               printf("%d",i);
           }
       printf("
");
    }
    return 0;
}