HDU 6092：Rikka with Subset

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：给一个数列a[1.....n],m是a数列所有元素的和,给出数列b[0.....m],b[i]代表a数列的子集中和为i的子集的个数。现在给出数列b，让你求数列a（有多解的话，字典序要最小，不过好像只有一个解。。。）。

思路：逆向DP。

可以想到b[i]是由a数列中的元素进行DP得到的。

即：

for(int i=1;i<=n;i++)

{

for(int j=m;j>=a[i];j--)b[j]+=b[j-a[i]];

}

那么我们就可以反过来求出数列a；

#include<bits/stdc++.h>using namespace std;const int maxn = 1e4+5;long long b[maxn],a[100];int main(){long long T,n,m;cin>>T;while(T--){cin>>n>>m;for(long long i=0;i<=m;i++)scanf("%lld",&b[i]);for(long long i=1;i<=m;i++){if(b[i]){a[1]=i;break;} //求出a的第一个数（因为最小的i且b[i]!=0一定代表i是a中的最小数）}long long cnt=2;for(long long i=1;i<=n;i++) //从第一个数反过来DP{for(long long j=a[i],tag=0;j<=m;j++){b[j]-=b[j-a[i]];if(b[j]>0&&tag==0)a[cnt++]=j,tag=1; //每次都要把最小的i且b[i]!=0的数加入a中}}for(long long i=1;i<=n;i++)printf("%lld%c",a[i],i==n?'\n':' ');}return 0;}