2017 HDU 6092 多校联合赛 Rikka with Subset

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 nn 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≤10e4)

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]

题意：

有两个数组a[ ]和b[ ]，给出两个数n和m，表示a[ ]中有 n 个数，所有数之和为 m ，然后又给出了b[ ]数组，

b[ x ]数组中装的是a[ ]中和为 x 的子集个数。例如：a[ ]数组为[ 1，2 ]；那么它的子集有[ ]，[ 1 ]，[ 2 ]，

[ 1，2 ]；这些子集的和分别为 0，1，2，3；所以b[ 0 ]=1，b[ 1 ]=1，b[ 2 ]=1，b[ 3 ]=1；现在给你b[

]数组，让你求a[ ]数组。

思路：

一个很好的DP题目，以前做过一个用零钱凑总共钱数的问题，这个题目反向，由总钱数求零钱。dp[ x ]中

装的是由小数加和而成的 x 有多少个，然后b[ x ]-dp[ x ]就是真正有多少个 x，然后在再用求出的 x 个数与

前面的小数继续组合求和，不断更新dp[ ]。

代码：

#include<cstdio>#include<cstring>using namespace std;const int Max = 10e4+10;int dp[Max];int a[55];int b[Max];int num[Max];int t,n,m;void work(){    dp[0]=1;    int ant=1;    for(int i=1;i<=m;i++)    {        num[i]=b[i]-dp[i];        for(int u=1;u<=num[i];u++)        {            a[ant++]=i;            for(int j=m;j>=i;j--)            {                dp[j]=dp[j]+dp[j-i];            }        }    }}int main(){    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(num,0,sizeof(num));        scanf("%d%d",&n,&m);        for(int i=0;i<m+1;i++)        {            scanf("%d",&b[i]);        }        work();        for(int i=1;i<=n;i++)        {            printf("%d%c",a[i],i==n?'\n':' ');        }    }    return 0;}