[HDU 6092] Rikka with Subset

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 139    Accepted Submission(s): 49

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 $B=[1,1,1,1]$
题意：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
思路： 
分析：很多个较小的数字随机组合会求出多个很大的数字，所以从B0向Bm推导，在每求出A序列的一部分这个过程中，更新后续的B序列，更新完的B[i]就是 i 在A序列中出现的次数。 
分析完后，主要的难点就是怎么去让已求出来的A序列随机组合，更新后续的B序列直接减就可以了。看成01背包问题，让m为背包去装 i，初始值为dp[0] = 1，由于i依次增大，A子集随机组合不会重复
如果 B_iB​i​​ 是 $B$ 数组中除了 B_0B​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 B_j-=B_{j-i}B​j​​−=B​j−i​​。
时间复杂度 $O(nm)$。

AC代码

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define MAX 11000int main(){int T;int n,m;int B[MAX],dp[MAX],num[MAX],ans[MAX];//dp[i]表示：加和为i的子集个数；数组存的是用比i小的数能组成i的情况scanf("%d",&T);while(T--){scanf("%d %d",&n,&m);for(int i=0;i<=m;i++){scanf("%d",&B[i]);}memset(ans,0,sizeof(ans));memset(num,0,sizeof(num));memset(dp,0,sizeof(dp));dp[0]=1;//空集的情况 int cnt=0;for(int i=1;i<=m;i++){num[i]=B[i]-dp[i];//A序列中值为i的个数if(cnt==n) break;for(int j=1;j<=num[i];j++){ans[cnt++]=i; //对A序列赋值for(int k=m;k>=i;k--)//处理成01背包 dp[k]+=dp[k-i];//和为k的A子集个数相加去更新B序列}}for(int i=0;i<n;i++)printf("%d%c",ans[i],i==n-1?'\n':' ');}return 0;}

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 139    Accepted Submission(s): 49

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