HDU6092-Rikka with Subset

Rikka with Subset

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                             Total Submission(s): 1535    Accepted Submission(s): 775

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  i s 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

 

题意：有一个长度为n的数组a，告诉你数组b，b[i]表示元素和为i的集合的个数，求出数组a，并按字典序输出

解题思路：首先题目中0的个数和1的个数必然是确定的，根据1的个数，我们可以算出它对后面数字的贡献，然后后面这些数字减去这个贡献，到最后这些数字就是自己对自己的贡献，即每种数字有几个

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, m;
LL c[55][55], b[10011];

void C()
{
    int i, j;
    c[0][0] = 1;
    for(i = 1; i < 55; i++)
    {
        c[i][0] = c[i][i] = 1;
        for(j = 1; j < i; j++)
        {
            c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
        }
    }
}

int main()
{
    C();
    int t;
    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] == 0) continue;
            for(int j = i + 1; j <= m; j++)
            {
                for(int k = 1; k <= b[i]; k++)
                {
                    if(j - i * k < i) break;
                    if(j - i * k != i) b[j] -= c[b[i]][k] * b[j - i * k];
                    else b[j] -= c[b[i]][k + 1];
                }
            }
        }
        int flag=1;
        for(int i = 1; i <= m; i++)
        {
            while(b[i]--)
            {
                if(flag) flag=0;
                else printf(" ");
                printf("%d", i);
            }
        }
        printf("\n");
    }
    return 0;
}