HDU 6092 Rikka with Subset 【dp多重背包】【好题】

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 139    Accepted Submission(s): 49

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：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

显然数字0的数量num[0]为log2(b[0])，数字1的数量num[1]为b[1]/b[0]

设dp[i]，表示在当前i没有的情况下，用前面已知数量的数组成数字i共有多少种情况

那么b[i]-dp[i]即为数字i与0进行组合的可能性，则num[i]=(b[i]-dp[i])/b[0]

这里如果直接写的话复杂度为o(m^2)会超时，所以需要剪枝：

if (dp[j] == 0) continue;
if (num[i] == 0) break;

直接将m^2的复杂度降为nm

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#define ms(a,b) memset(a,b,sizeof(a)) 
using namespace std;
typedef long long ll;

const int maxn = 1e4 + 100;

ll b[maxn];
int dp[maxn], num[maxn];

int C(int n, int m)
{
	int sum = 1;
	for (int i = n - m + 1; i <= n; i++) sum *= i;
	for (int i = 1; i <= m; i++) sum /= i;
	return sum;
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n, m;
		scanf("%d%d", &n, &m);
		ms(dp, 0);
		ms(num, 0);
		for (int i = 0; i <= m; i++)
		{
			scanf("%lld", &b[i]);
		}
		num[0] = log2(b[0]);
		num[1] = b[1] / b[0];
		dp[0] = b[0];
		for (int i = 0; i <= m; i++)
		{
			for (int j = m; j >= 0; j--)
			{
				if (dp[j] == 0) continue;
				if (num[i] == 0) break;
				for (int k = 1; k <= num[i]; k++)
				{
					if (j + k*i <= m)
					{
						dp[j + k*i] += dp[j] * C(num[i], k);
					}
				}
			}
			if (i + 1 <= m)
			{
				num[i + 1] = (b[i + 1] - dp[i + 1]) / b[0];
			}
		}
		bool flag = 0;
		for (int i = 0; i <= m; i++)
		{
			for (int j = 1; j <= num[i]; j++)
			{
				if (!flag) printf("%d", i), flag = 1;
				else printf(" %d", i);
			}
		}
		puts("");
	}
	return 0;
}