Learn Python from a young age! Record the questions in the Luogu Python learning and exam preparation process, and record every moment.
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[Title description]
Teacher P needs to go to the store to buy n pencils as gifts for children participating in NOIP. She finds that there are 3 packages of pencils in the store. The number of pencils in different packages may be different, and the price may also be different. To be fair, Teacher P decided to only buy pencils in the same package.
The store does not allow the packaging of pencils to be opened, so Teacher P may need to buy more than n pencils to give gifts to the children.
Now Teacher P wants to know how much it costs to buy at least n pencils if the store has enough of each package.
【enter】
The first line contains a positive integer n, representing the number of pencils required.
The next three lines each use 2 positive integers to describe a package of pencils: the first integer represents the number of pencils in this package, and the second integer represents the price of this package.
Ensure that all 7 numbers are positive integers not exceeding 10,000.
【Output】
1 integer, indicating the minimum amount of money that Teacher P needs to spend.
【Input sample】
57 2 2 50 30 30 27
【Output sample】
54
[Detailed code explanation]
n = int(input())
c1, p1 = [int(i) for i in input().split()]
c2, p2 = [int(i) for i in input().split()]
c3, p3 = [int(i) for i in input().split()]
# 计算第一组铅笔需要花费的金额
if n % c1 == 0:
pp = n // c1 * p1
else:
pp = (n // c1 + 1) * p1
minn = pp
# 计算第二组铅笔需要花费的金额
if n % c2 == 0:
pp = n // c2 * p2
else:
pp = (n // c2 + 1) * p2
if minn > pp:
minn = pp
# 计算第二组铅笔需要花费的金额
if n % c3 == 0:
pp = n // c3 * p3
else:
pp = (n // c3 + 1) * p3
if minn > pp:
minn = pp
print(minn)【operation result】
9998
128 233
128 2333
128 666
18407