HDU 6092 Rikka with Subset（背包+思维）-优快云博客

本文链接：https://blog.youkuaiyun.com/Expecto__Patronum/article/details/76992444

本文介绍了一个数学问题，通过给定的序列B，其中包含了序列A所有子集的和的出现次数，目标是找出字典序最小的序列A。文章提供了解决这一问题的方法，包括确定序列A中的最小数及其后序列B的更新过程。

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Rikka with Subset

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

题意：给出序列a的长度n，序列元素和为m，给出序列b，其中bi为序列a任意子集合sum为i的个数，输出字典序最小的序列a

题解：如果 $B_i$ 是 $B$ 数组中除了 $B_0$ 以外第一个值 $0$ 的位置，那么显然 $i$ 就是 $A$ 中的最小数。

现在需要求出删掉 $i$ 后的 $B$ 数组，过程大概是反向的背包，即从小到大让 $B_j-=B_{j-i}$ 。

时间复杂度 $O (n m)$ 。

#include<iostream>
#include<stdio.h>
#include<string.h>
#define ll long long
using namespace std;
int b[10005];
int a[55];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0;i<=m;i++)
            scanf("%d",&b[i]);
        int coun=0;
        while(coun<n)
        {
            int i=1;
            for(i;i<=m;i++)
            {
                if(b[i]!=0)
                    break;
            }
            b[i]--;
            a[coun++]=i;
            for(int j=i+1;j<=m;j++)
                b[j]-=b[j-i];
        }
        for(int i=0;i<n;i++)
        {
            cout<<a[i];
            if(i==n-1) cout<<endl;
            else cout<<" ";
        }
    }
    return  0;
}