Arrays HackerRank产品分销_Arrays_Python 3.x_Algorithm_Functional Programming

Arrays HackerRank产品分销

arrays python-3.x algorithm functional-programming

Arrays HackerRank产品分销,arrays,python-3.x,algorithm,functional-programming,Arrays,Python 3.x,Algorithm,Functional Programming,我一直在努力解决这个挑战，并在一定程度上想出了该怎么做。但是在我所有的尝试之后，我只通过了测试用例，并且在提交面板中又通过了一个用例。之后一切都失败了问题： def maxScore(segment, products): # Write your code here # If the segment == products, then it should return all the sum # We will evaluate as per the products

我一直在努力解决这个挑战，并在一定程度上想出了该怎么做。但是在我所有的尝试之后，我只通过了测试用例，并且在提交面板中又通过了一个用例。之后一切都失败了

问题：

def maxScore(segment, products):
    # Write your code here
    # If the segment == products, then it should return all the sum
    # We will evaluate as per the products listing requirement and find the sum

    '''
        Algo for else condition
        1. We will maintain a start and end pointer to keep a check till counter equals products
        2. We will keep adding the maxSum of the value [i * sum(batch of the element)]
        3. Come out of the loop and perform a final operation as above with the remaining elements
        4. Add it to sum
        5. Return maxSum
    '''
    batch = 1
    maxSum = 0
    start = 0 
    end = products
    segment.sort()
    while batch != products:
      maxSum += (batch * sumElem(segment[start:end]))
      batch += 1
      start += products
      end += products
    maxSum += (batch * sumElem(segment[start:len(segment)]))
    return maxSum

# function to find the sum of the elements
def sumElem(arr):
    total = 0
    for item in arr: total += item
    return total

一家公司要求简化其产品分配策略，并且给定n个产品，每个产品都有一个关联的值，您需要将这些产品分为多个部分进行处理。有无穷多段索引为1、2、3等等

但是，有两个限制：

当且仅当i=1或索引为i-1的段至少有m个产品时，可以将产品分配给索引为i的段
任何段必须不包含产品或至少包含m个产品

段的分数定义为段的指数乘以其包含的产品的值之和。产品排列的得分是各部分得分的总和。您的任务是计算安排的最大分数

例如，考虑n=11个产品和m=3。一种最佳的分配方式是-

将前三种产品分配给第1部分

将接下来的三种产品分配给第2部分

将接下来的五种产品分配给第3部分

请注意，我们不能将2个产品分配给段4，因为第二个约束将被违反。上述安排的分数为-

1*（1+2+3）+2*（4+5+6）+3*（7+8+9+10+11）=6+30+135=171

由于排列分数可能非常大，所以按10^9+7的模打印

输入格式

在第一行中，有两个空格分隔的整数n和m

在第二行中，有n个空格分隔的整数a0，a1，…，an-1表示与乘积相关的值

约束

1我们首先对给定数组进行排序。这是因为我们最终“分段”的数字越大，它将导致一个更大的总和，并且它将乘以分段指数，从而使输出最大化

解决这个问题的简单逻辑是找到

产品的最大段数。与示例输入一样，对于输入

n=5

，

m=2

：我们最多可以用2个产品组成1个细分市场，因为如果我们用2个产品组成2个细分市场，我们最终只剩下1个产品，而我们无法将其细分（根据问题）

为了达到这个目的，我们用

除以字符串的长度，如果它不能完全整除，则从中减去1

def maxScore(a, m):

a.sort()
print(a)
x=len(a)

if x%m==0:
    y=int(x/m)
else:
    y=int(x/m)-1

summ=0
count=1
#print(y)
i=0
for _ in range(y): 
    summ=summ+(sum(a[i:i+m])*count)
    count=count+1
    i=i+m
    print(summ)

summ=summ+sum(a[i:])*count
print(summ)

return summ%1000000007

def最大分数（a，m）：
#在这里编写代码
a、 排序（）
l=len（a）
j=1
s=0
i=0
而i+（2*m）也许你也可以分享问题链接。我找不到它。是来自正在进行的比赛吗？是的@shivank98，比赛名称是Hack the Interview II-Global，否则我会分享这个链接。不管怎样，这个问题写得很清楚是为了让你们理解。我不确定，但实际上，在这种情况下，你们试图在某处找到答案是一种欺骗。它似乎没有遵循这样的政策。有人提到我找不到链接。我没有作弊，因为我已经用我的代码解决了这个问题。只是我需要一些指导方面的帮助。那是allI，我看不出你在哪里取模。我对Python不太了解，但如果变量范围不够大，那么在末尾取模是不可行的
4 4
4 1 9 7

21

def maxScore(segment, products):
    # Write your code here
    # If the segment == products, then it should return all the sum
    # We will evaluate as per the products listing requirement and find the sum

    '''
        Algo for else condition
        1. We will maintain a start and end pointer to keep a check till counter equals products
        2. We will keep adding the maxSum of the value [i * sum(batch of the element)]
        3. Come out of the loop and perform a final operation as above with the remaining elements
        4. Add it to sum
        5. Return maxSum
    '''
    batch = 1
    maxSum = 0
    start = 0 
    end = products
    segment.sort()
    if len(segment) == products:
        maxSum += (batch * sumElem(segment[start:len(segment)]))
    else: 
        while batch != products:
            maxSum += (batch * sumElem(segment[start:end]))
            batch += 1
            start += products
            end += products
        maxSum += (batch * sumElem(segment[start:len(segment)]))
    return maxSum

# function to find the sum of the elements
def sumElem(arr):
    total = 0
    for item in arr: total += item
    return total

def maxScore(segment, products):
    # Write your code here
    # If the segment == products, then it should return all the sum
    # We will evaluate as per the products listing requirement and find the sum

    '''
        Algo for else condition
        1. We will maintain a start and end pointer to keep a check till counter equals products
        2. We will keep adding the maxSum of the value [i * sum(batch of the element)]
        3. Come out of the loop and perform a final operation as above with the remaining elements
        4. Add it to sum
        5. Return maxSum
    '''
    batch = 1
    maxSum = 0
    start = 0 
    end = products
    segment.sort()
    while batch != products:
      maxSum += (batch * sumElem(segment[start:end]))
      batch += 1
      start += products
      end += products
    maxSum += (batch * sumElem(segment[start:len(segment)]))
    return maxSum

# function to find the sum of the elements
def sumElem(arr):
    total = 0
    for item in arr: total += item
    return total

def maxScore(a, m):
    # Write your code here

    n=len(a)
    a.sort()
    possible_quotient=n//m
    sets=possible_quotient-1
    set_range=sets*m
    base=1
    SUM=0
    k=0
    for i in range(sets):
        temp=a[k:k+m]
        k+=m
        tsum=sum(temp)*base
        SUM+=tsum
        base+=1
        i+=m
    temp2=sum(a[set_range:n])*base
    SUM+=temp2
    return(SUM%1000000007)

def maxScore(a, m):

a.sort()
print(a)
x=len(a)

if x%m==0:
    y=int(x/m)
else:
    y=int(x/m)-1

summ=0
count=1
#print(y)
i=0
for _ in range(y): 
    summ=summ+(sum(a[i:i+m])*count)
    count=count+1
    i=i+m
    print(summ)

summ=summ+sum(a[i:])*count
print(summ)

return summ%1000000007

def maxScore(a, m):
    # Write your code here
    a.sort()
    l = len(a)
    j = 1
    s = 0
    i = 0
    while i+(2*m) <= l:
        s += sum(a[i:i+m])*j
        i += m
        j += 1
        print(s)
    s += sum(a[i:])*j
    return s%1000000007