HDU6092-Rikka with Subset

最新推荐文章于 2019-02-17 10:39:30 发布

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本文介绍了一个数学问题RikkawithSubset，该问题要求根据给定的子集和频次恢复原始集合。文章详细阐述了解题思路及算法实现过程，包括如何通过迭代计算确保解的正确性和唯一性。

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Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 513 Accepted Submission(s): 218

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

Source
2017 Multi-University Training Contest - Team 5

题目大意：已知集合 $A$ 有n个元素（正整数），元素和为 $m$ ，对于集合 $A$ 的 $2^n$ 个子集，分别计算出它们元素的和，得到 $2^n$ 个数，对于 $i\in [0,m]$ ，记 $B_i$ 为这 $2^n$ 个数中 $i$ 的个数。给出 $B$ 集合，求字典序最小的 $A$ 集合。
解题思路： $B_0=1$ （空集）,那么遍历 $i$ 从 $1$ 到 $m$ ，若 $B_i >0$ ，则将 $i$ push_back进vector,对于 $j\in[i,m]$ ， $j$ 可以由当前数 $i$ 加上能够组合成和为 $j-i$ 的集合构成，那么只要将 $B_j$ 减去 $B_{j-i}$ ，就能得到在没有 $i$ （单独这一个 $i$ ）的情况下能够组合成和为 $j$ 的情况数，从小到大遍历保证了结果正确性。重复直至得到vector中有 $n$ 个数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAXN=1e4+5;
LL b[MAXN];
vector<int> v;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        v.clear();
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<=m;i++)
            scanf("%lld",&b[i]);
        for(int i=1;i<=m;++i)
        {
            if(v.size()==n) break;
            while(b[i]>0)
            {
                v.push_back(i);
                for(int j=i;j<=m;++j)
                {
                    b[j]-=b[j-i];
                }
            }
        }
        bool flag=true;
        for(auto i:v)
        {
            if(!flag) printf(" %d",i);
            else {printf("%d",i);flag=false;}
        }
        printf("\n");
    }
    return 0;
}