从给定数组派生MultiSet，使得sum为> P，除去任何元素使sum <P

给定N个元素的数组arr [] ，任务是从给定数组中以可能的重复数得出具有数字的MultiSet ，以使MultiSet的总和严格大于给定的数P，并且如果删除了任何元素，则总和严格小于P。打印给定数组的对应元素派生到MultiSet中的次数。如果无法导出此类MultiSet，则打印-1。

例子：

Input: arr[] = [1, 5], P = 4
Output: [0, 1] 
Explanation: 
Here, if number 1 is taken 0 times, and 5 is taken 1 time, the MultiSet becomes: [5]
Therefore, sum = 5 (>P) and removing 5 makes sum = 0 (Therefore, the output is [0, 1] as the number of times 1 and 5 are taken respectively.

Input: arr[] = [1, 5], P = 10
Output: -1
Explanation: 
If we take a multiset as [1, 5, 5], the sum will be > P, but removing 1 will not make sum < P. Hence, no such MultiSet can be derived in this case.

方法：
上面问题的主要观察结果是，如果P被数组arr []中的任何元素不可分割，则我们采用该元素的所有倍数，使得总和严格大于p。但是，如果数组中没有这样的元素，并且有可能使用多集，那么我们将以降序对数组进行排序，并取每个元素的倍数小于P％arr [i]并保持更新P。多集是不可能的如果每次更新的P都可被arr [i + 1]整除到N。下面是上述方法的实现：

Python3

# Python implementation to Derive a
# MultiSet from given Array such that
# sum is > P and removing any
# element makes sum < P
  
# Function to derive the multiset
def Multiset (n, p, arr):
      
    c = 0
      
    for j in arr:
          
        # Check if p is indivisible
        # by any element in arr
        if (p % j != 0):
            c = j
            break
              
    # Check if there is no element in
    # arr which cannot divide p    
    if (c == 0):
          
        d = sorted(arr)
          
        # Sort arr in descending order
        d = d[::-1]         
        coun = 0
        pri = [0] * n
          
        # Assigning multiples of each element
        while (coun != n and p % d[coun] == 0): 
            
            b = arr.index(d[coun])
            pri[b] = ((p//d[coun]) - 1)
            p = p - (d[coun]*((p//d[coun]) - 1))
            coun += 1
          
        # Check if there is no case
        # of getting sum > p
        if (coun == n):
            return ("NO")
              
        elif (p % d[coun] != 0):
            y = (p//d[coun]) + 1
            k = arr.index(d[coun])
              
            pri[k] = y
            s = ""
              
            # Multi set
            for j in pri:                     
                s = s + str(j) + " "
            return (s)
          
              
    else:
        k = p//c
        b = c * (k + 1)
        m = [0] * n
        q = arr.index(c)
        m[q] = b//c
        s = ""
        for j in m:
            s = s + str(j) + " "
        return (s)
  
# Driver code
N, P = 2, 4
arr = [1, 5]
print (Multiset(N, P, arr))

输出

0 1 

时间复杂度： O(N)
辅助空间： O(1)