HDU6092 Rikka with Subset-01背包dp-2017多校联盟5 第8题

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

22 31 1 1 13 31 3 3 1

 

Sample Output

1 21 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：一个序列a有n个数,和为m，这个序列能组成2^n个子序列，bi表示这些子序列中和为i的个数现在给你bi，i∈[0,m]求出序列a，a按字典序输出解题思路首先能明确的是b[0]一定等于1，并且b中第一个不为0的i一定有b[i]个。b[i]=i的个数+组合成i的方案数.当确定a中存在i，并且能组合出j-i，那么一定能组合出j,这种途径得到j的方案数就是j-i的方案数*/#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 1e4 + 10;int dp[maxn];///组合成i的方案数，不包括i本身,（准确的说是在完成第46行循环之前不包括本身。）int b[maxn];int n,m;int main(){    int t;    scanf("%d", &t);    while (t > 0)    {        t--;        memset(dp,0,sizeof(dp));        dp[0] = 1;        scanf("%d%d",&n,&m);        for(int i = 0;i<=m;++i){            scanf("%d",&b[i]);        }        bool f = 0;        for(int i = 1;i<=m;++i){            int cnt = b[i]-dp[i];///b[i]=i的个数+组成i的方案数dp[i]            if(cnt<=0) continue;///不存在i            for(int k = 0;k<cnt;++k){///a里面肯定有cnt个i                if(f) printf(" ");                f=1;                printf("%d",i);                for(int j = m;j>=i;--j){                    dp[j] += dp[j-i];///组成j-i的方案分别加上i就能组成j，加上这种途径获得的方案数                    //dp[j+i] += dp[j];一样的也能过,注意j的范围                }            }        }        puts("");    }    return 0;}