2017 Multi-University Training Contest

题目：

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 numbert(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 withn 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：给你一个B数组，共m+1个数，求A数组。B[i] 代表A序列中的所有子集之和为i的有B[i]个，A序列总和为m，共n个元素。

思路：大的数都是由很多个小的数组成，从小到大枚举 i，然后判断我们需要用几个i，用dp数组来维护此时共凑到多少个i，不够的（t>0）我们就只能直接添加i，接着在每求出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......发现每次做动态规划、dp的问题，总能被自己蠢哭，脑子不够用~

code：

#include<bits/stdc++.h>using namespace std;__int64 a[10005],b[10005],dp[10005];   //dp[i]表示加和为i的子集个数int main(){        int t,n,m,i,j,k;        scanf("%d",&t);        while(t--){                scanf("%d%d",&n,&m);                for(i=0;i<=m;i++) scanf("%I64d",&b[i]);                memset(dp,0,sizeof(dp));                k=1;dp[0]=1;   //初始化值                for(i=1;i<=m;i++){                        int t=b[i]-dp[i];   //A序列中值为i的个数=和为i的子集个数-已经凑到的和为i的个数                        while(t--){                                a[k++]=i;                                for(j=m;j>=i;j--) dp[j]+=dp[j-i];   //和为j的子集个数相加更新B序列                        }                }                for(i=1;i<=n;i++) printf("%I64d%c",a[i],i==n?'\n':' ');        }        return 0;}