HDU6092 Rikka with Subset

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 403    Accepted Submission(s): 173

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

 
题目大意：给你一个n和m下一行输入B数列从B0~Bm，下标为i则Bi的含义为A数列的子集中和为i的子集为Bi个，现在让你还原A数列，并按照字典顺序输出。

解题思路：B0一定为零不用考虑，那我们现在从B1开始遍历遇到Bi不为零的时候比如B1不为零那么在A数列中1的个数就为B1个，假如B1=0，B2不为零那么同理从B2
开始。我们将这Bi个i一个个的拿走每拿走一个就计算一下拿走这一个i后B数列的值为多少，现在这个状态我们视为一种新的状态，再进行一次这个操作，知道把A数列全部拿完为止。就是一步步的将A拿出来，这样也正好是按照的字典顺序。
用这个解释一下样例，3 3     1 3 3 1

B1=3，即3个1（其实现在就是最后的答案了）那么我们拿一个1，拿了这一个1对原数列的影响：1 2 1 0即用Bj=Bj-B[j-i],B1-1=2,B2-B1=1,B3-B2=0.现在我们用答案验证是符合的。这样一次次的拿将A数列中的数拿完。

AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int t,n,m;long long b[10050]={0};int a[600]={0};int main(){scanf("%d",&t);while(t--){memset(b,0,sizeof(b));memset(a,0,sizeof(a));scanf("%d%d",&n,&m);for(int i=0;i<=m;i++){scanf("%lld",&b[i]);}int k=0;for(int i=1;i<=m;i++){if(k>=n){break;}if(i<=0){continue;}if(b[i]!=0){    a[k]=i;k++;for(int j=i;j<=m;j++){b[j]=b[j]-b[j-i];}i--;}}printf("%d",a[0]);for(int i=1;i<k;i++){printf(" %d",a[i]);}printf("\n");}return 0;}