Rikka with Subset（hdu6092）

最新推荐文章于 2017-08-21 21:02:38 发布

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本文介绍了一个数学问题，即根据子集和频次反推原始数组的过程。通过输入子集和频次数组，解决如何找到原始数组的问题，并给出了具体的实现代码。

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

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：一共有T组测试样例，然后n和m，有m+1个数代表B数列，表示为A数列子集的和为i的有多少种；
解题思路：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

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
int n,m;
int ans[10005],tmp[10005],dp[10005];
int da[10005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {


        scanf("%d %d",&n,&m);
        for(int i=0;i<=m;i++)
        scanf("%d",&da[i]);
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));
        memset(tmp,0,sizeof(tmp));
        dp[0]=1;
        int p=0;
        for(int i=1;i<=m;i++)
        {
            tmp[i]=da[i]-dp[i];
            for(int j=0;j<tmp[i];j++)
            {
                ans[p]=i;
                p++;
                for(int k=m;k>=i;k--)
                {
                    dp[k]+=dp[k-i];//和为k的子集不断更新
                }
            }
        }
        for(int i=0;i<n-1;i++)
        printf("%d ",ans[i]);
        printf("%d\n",ans[n-1]);
    }
    return 0;
}