A new variety of rice has been brought in supermarket and being available for the first time, the quantity of this rice is limited. Each customer demands the rice in two different packaging of size a and size b. The sizes a and b are decided by staff as per the demand.
Given the size of the packets a and b, the total quantity of rice available d and the number of customers n, find out maximum number of customers that can be satisfied with the given quantity of rice.
Display the total number of customers that can be satisfied and the index of customers that can be satisfied.
Note: If a customer orders 2 3, he requires 2 packets of size a and 3 packets of size b. Assume indexing of customers starts from 1.
The first line of input contains two integers n and d; next line contains two integers a and b. Next n lines contain two integers for each customer denoting total number of bags of size a and size b that customer requires.
Print maximum number of customers that can be satisfied and in next line print the space separated indexes of satisfied customers.
Input : n = 5, d = 5 a = 1, b = 1 2 0 3 2 4 4 10 0 0 1 Output : 2 5 1 Input : n = 6, d = 1000000000 a = 9999, b = 10000 10000 9998 10000 10000 10000 10000 70000 70000 10000 10000 10000 10000 Output : 5 1 2 3 5 6
In first example, the order of customers according to their demand is:
Customer ID Demand 5 1 1 2 2 5 3 8 4 10
From this, it can easily be concluded that only customer 5 and customer 1 can be satisfied for total demand of 1 + 2 = 3. Rest of the customer cannot purchase the remaining rice, as their demand is greater than available amount.
In order to meet the demand of maximum number of customers we must start with the customer with minimum demand so that we have maximum amount of rice left to satisfy remaining customers. Therefore, sort the customers according to the increasing order of demand so that maximum number of customers can be satisfied.
Below is the implementation of above approach:
# Python3 program to find maximum number
# of customers that can be satisfied
v = 
# print maximum number of satisfied
# customers and their indexes
def solve(n, d, a, b, arr):
first, second = 0, 1
# Creating an vector of pair of
# total demand and customer number
for i in range(n):
m = arr[i]
t = arr[i]
v.append([a * m + b * t, i + 1])
# Sorting the customers according
# to their total demand
ans = 
# Taking the first k customers that
# can be satisfied by total amount d
for i in range(n):
if v[i][first] <= d: ans.append(v[i][second]) d -= v[i][first] print(len(ans)) for i in range(len(ans)): print(ans[i], end = " ") # Driver Code if __name__ == '__main__': # Initializing variables n = 5 d = 5 a = 1 b = 1 arr = [[2, 0], [3, 2], [4, 4], [10, 0], [0, 1]] solve(n, d, a, b, arr) # This code is contributed by PranchalK [tabbyending] Output:
2 5 1
- Divide N segments into two non-empty groups such that given condition is satisfied
- Maximum number of contiguous array elements with same number of set bits
- Find the maximum number of composite summands of a number
- Find row number of a binary matrix having maximum number of 1s
- Find the row with maximum number of 1s
- Maximum number of candies that can be bought
- Maximum number of ones in a N*N matrix with given constraints
- Maximum number of segments that can contain the given points
- Level with maximum number of nodes using DFS in a N-ary tree
- Possible cuts of a number such that maximum parts are divisible by 3
- Maximum number of teams that can be formed with given persons
- Find the Number of Maximum Product Quadruples
- Python map function to find row with maximum number of 1's
- Paths with maximum number of 'a' from (1, 1) to (X, Y) vertically or horizontally
- Find sum of a number and its maximum prime factor
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : PranchalK