HDU
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Rikka with Subset
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 412 Accepted Submission(s): 178
Problem DescriptionAs 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 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?
InputThe 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) .
OutputFor 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 Input22 31 1 1 13 31 3 3 1
Sample Output1 21 1 1HintIn 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 $B=[1,1,1,1]$
Source2017 Multi-University Training Contest - Team 5
Yuta has
Now, Yuta has got
Yuta shows Rikka the array
It is too difficult for Rikka. Can you help her?
For each testcase, the first line contains two numbers
The second line contains
It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.
22 31 1 1 13 31 3 3 1
1 21 1 1HintIn 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 $B=[1,1,1,1]$
题目大意:
有一个数列 a[] ,长度(n<=50)。b[i] 表示元素和为 i 的集合个数。给你一个数列 b[] ,长度(m<=10000),让你求 a[],并按照其字典序最小输出。
分析:
首先,对于除了 b0 以外的第一个不为 0 的 bi ,数组 a[] 里面一定有 i,并且有 b[i]个(但是我不是一次就把这 b[i] 个一次性全拿出来,我一个一个拿) 。首先进行完拿出一个的操作之后,然后要对 b 进行操作,把 b 数组变成去除 i 后继续满足原定义。首先考虑,对于每一个和为 j 的组合(元素中没有 i ),把它里面加上一个元素 i 它就变成了一个和为 j+i 的组合。
也就是说:
所以按照这种方法就可以在:O(mn) 的时间内得到 n 个数,然后按照字典序最小输出即可。
注意尾部不能有空格。
代码:
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