举例来说,我想给一些员工一些钱。每个员工必须得到 $a 到 $b 美元。第一个员工得到 $a,每个后续员工得到的 $k 比最后一个多,直到该金额超过 $b,在这种情况下,该员工获得 $b,此后每个后续员工得到的 $k 比最后一个少,直到该金额低于 $b $a 在这种情况下,员工将获得 $a 并且所有 n 名员工都将继续循环。我想将总支出返还给所有员工
我目前拥有的:
#!/bin/python3
import os
import sys
def payEmp(n, a, b, k):
totalPayOut = 0
currentPay = 0
increase = True
for i in range(n):
if increase == True:
if currentPay < a:
currentPay += a
else:
currentPay += k
if currentPay >= b:
totalPayOut += b
increase = False
else:
totalPayOut += currentPay
else:
currentPay -= k
if currentPay <= a:
totalPayOut += a
increase = True
else:
totalPayOut += currentPay
return totalPayOut
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
t = int(input())
for t_itr in range(t):
nabk = input().split()
n = int(nabk[0])
a = int(nabk[1])
b = int(nabk[2])
k = int(nabk[3])
result = payEmp(n, a, b, k)
fptr.write(str(result) + '\n')
fptr.close()
最佳答案
我可能会使用生成器函数来创建支出:
def pay(n,a,b,k):
p = a # start with a
c = 0
while c < n: # loop until enough values generated
# upcounting payments
while p <= b and c < n: # do this until enough or payment reached b
yield p
c += 1
if p != b:
p = min(b,p+k) # increase as long as not reached, prevent overshooting
else: # we reached and yielded b so we are done
break # we just yielded b - less payment from now on
p -= k # we already yielded b - so we add k again
# downcounting payments
while p >= a and c < n: # do this until enough or payment reached a again
yield p
c += 1
if p != a:
p = max(a,p-k) # decrease as long as not reached, prevent undershooting
else: # we just yielded a, were done going down, back up from now on
p = a+k
break # we just printed a, more pay from here on
pays = list(pay(15,2,9,2))
print(pays,sum(pays))
输出:
[2, 4, 6, 8, 9, 7, 5, 3, 2, 4, 6, 8, 9, 7, 5] 85
关于python - 支付员工,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50009692/