Question

我正在编写一个学费程序。我的问题/问题是：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
'''

Answer 1

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()

每笔学费都会追加（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。

然后我在字符串中使用内联变量，只是因为这就是我喜欢这样做的方式。我对列表求和，再次将其限制为前4个项目，然后将其限制为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个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()

Answer 2

您可以创建一个列表来保存每年的值。在此示例中，列表的元素“ i”对应于年份“ i + 1”，因为列表从0变为4

tuition = 8000.00
tuition_year = [tuition]
for i in range(1, 5):
    tuition_year.append(tuition_year[i-1] * 1.03)

total = sum(tuition_year)

Answer 3

def tuition(base, inc, size):
    for x in range(size):
        base += base * inc
        yield base

print("Year\t", "Tuition per semester")
for i, x in eunmerate(tuition(8000, 0.03, 5)):
    print(i+1, "\t", format(x, ".2f"))

total = sum(tuition(8000, 0.03, 4))
print("Total tuition in 4 years: {.2f}".format(total))

Answer 4

还要在for循环之前创建一个empty list，用于存储学费，然后调用python sum()功能，以返回列表中所有值的总和。

给出的代码

    tutions=[]
    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'))
        tutions.append(tutions)

    total += sum(tutions)
    print("Total tuition in 4 years: $", format(total,\
          '.2f'),sep = '')

每学期学费计划

4 个答案: