我正在写一个学费计划。我的问题/问题是：我如何才能获得四年的总数，而不是像我一样写出总数+ =（8240 + 8487.20 + 8741.82 + 9004.07）* 2
'''
CPS 313: Assignment 3, problem 3
At one college, the tuition for a full time student
is $8,000 per semester. It has been announced that the
tuition will be increased by 3% each year for the next 5 years.
Write a program with a loop that displays the projected semester
tuition amount for the next 5 years, and the total cost of 4
years' worth of tuition starting from now.
'''
def main():
tuition = 8000.00
total = 0
print("Year\t", "Tuition per semester")
for i in range(1,6):
#Tuition increased by 3% each year for the next 5 years
tuition += tuition*0.03
#Displays tuition fee for each year
print(i,'\t',format(tuition, '.2f'))
total += (8240+8487.20+8741.82+9004.07)*2
print("Total tuition in 4 years: $", format(total,\
'.2f'),sep = '')
main()
'''
Output:
Year Tuition per semester
1 8240.00
2 8487.20
3 8741.82
4 9004.07
5 9274.19
Total tuition in 4 years: $68946.18
'''
最佳答案
def main():
tuition = 8000.00
tuition_list = []
total = 0
for i in range(5):
tuition += tuition*0.03
tuition_list.append(tuition)
for i, x in enumerate(tuition_list[:4]):
print(f'Year {i+1}\tTuition ${round(x)}')
print(f'Total for 4 years ${round(sum(tuition_list[:4]) * 2)}')
main()
每个学费都会被追加（添加）到列表中，因此最终会有5个元素的列表。
然后，我使用enumerate（）遍历列表。这使我可以访问每个列表项，还可以创建一个索引号（0-5）。我使用list [：4]将列表限制为4。
然后，我只是在字符串中使用内联变量，因为这就是我喜欢的方式。我对列表求和，再次将其限制为前4个项目，然后将其乘以2。
输出：
>>> Year 1 Tuition $8240
>>> Year 2 Tuition $8487
>>> Year 3 Tuition $8742
>>> Year 4 Tuition $9004
>>> Total for 4 years $68946
编辑：
针对OP的评论。
您可以将4学费分配给像这样的变量
def main():
tuition = 8000.00
tuition_list = []
tuition_yr1 = tuition * 1.03
tuition_yr2 = tuition_yr1 * 1.03
tuition_yr3 = tuition_yr2 * 1.03
tuition_yr4 = tuition_yr3 * 1.03
total_cost_over_4yrs = (tuition_yr1 + tuition_yr2 + tuition_yr3 + tuition_yr4) * 2
print('Year 1', tuition_yr1)
print('Year 2', tuition_yr2)
print('Year 3', tuition_yr3)
print('Year 4', tuition_yr4)
print('Total over 4 years', total_cost_over_4yrs)
main()