Question

我正在编写一个读取整数的程序＆＃34; n＆＃34;从用户，然后将数字1^2-2^2+3^2-4^2+...±n^2加在一起。例如，如果n = 7，那么程序将返回28.我已经这样做了：

n = int(input("n = "))
summ = 0

for number in range (1,n +1):
    square = (number**2)*((-1)**(number+1))
    summ += square
    print(square)

print("The loop ran",number,"times, the sum is", summ)

问题是我希望程序在总和到达输入之前结束＆＃34; k＆＃34;来自用户。

n = int(input("n = ")) k = int(input("k = ")) summ = 0 for number in range (1,n +1): square = (number**2)*((-1)**(number+1)) summ += square print(square) if summ > k: break print("The loop ran",number,"times, the sum is", summ)

如果k = 6，程序返回＆＃34;循环运行5次，总和为15＆＃34;，但15显然超过6.正确的答案是＆＃34;循环运行4次，总和是-10＆＃34;。有谁知道如何解决这一问题？我也尝试将if语句放在＆＃34;对于数字＆＃34;但是，返回＆＃34;循环运行了6次，总和为15＆＃34;。

Answer 1

您正在更新summ值，然后检查条件。您应该在实际将数字添加到总和之前检查总数。

for number in range(1, n+1):
    square = (number**2)*((-1)**(number+1))
    if summ + square > k:
        break
    summ += square
    ...

# this should work, assuming the rest of your code works.

Answer 2

在将广场添加到汇总

之前设置一个条件

if summ + square > 7:
    break

n = int(input("n = "))
summ = 0

for number in range (1,n +1):
    square = (number**2)*((-1)**(number+1))
    if summ + square > 7:
        break
    summ += square

    print(square)

print("The loop ran",number,"times, the sum is", summ)

结束for循环，Python

2 个答案: