HDU6092 Rikka with Subset(01背包，2017 HDU多校联赛 第5场)

题目：

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 972    Accepted Submission(s): 476


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

 

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思路：
有两个数组a[]和b[]，每一个b[i]代表在a数组里面一些元素的和加起来等于i的子集的个数。

题目给了一个n代表a数组的个数，给了一个m代表a数组中所有元素的总和

问题：求出a数组，并且输出。

定义dp[i]:当前和为i的子集有多少个

要和b[i]区分一下，b[i]是最终的结果，而dp[i]是当前正在推算的结果。

所以，令 X=b[i]-dp[i]，代表了最终状态的子集个数减去当前个数的子集个数，也就是需要添加X个和为i的子集，那么我们直接在a数组里添加X个i,就可以使dp[i]达到b[i]的状态,

然后我们再更新dp[k]的值，也就是dp[k]=dp[k]+dp[k-i]

代码：

#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define N 300
#define ll long long
using namespace std;
int a[10010],b[10010],dp[10010];
int main()
{
int t,n,m,x;
scanf("%d",&t);
while(t--)
{
mem(dp,0);
scanf("%d%d",&n,&m);
int p=0;
for(int i=0; i<=m; i++)
scanf("%d",&b[i]);
dp[0]=1;
for(int i=1; i<=m; i++)
{
int x=b[i]-dp[i];
for(int j=0; j<x; j++)
{
a[p++]=i;
for(int k=m; k>=i; k--)
dp[k]+=dp[k-i];
}
}
for(int i=0; i<p; i++)
{
if(i)printf(" ");
printf("%d",a[i]);
}
puts("");
}
return 0;
}