```python def reciprocal_sum(precision): sum = 0 i = 1 while True: term = 1/i if abs(term) < precision: break sum += (-1)(i+1) * term i += 2 result = 4 * sum return result precision = float(input("请输入精度:")) result = reciprocal_sum(precision) print("正负交替数列和的4倍结果为:", result) ``` 你可以尝试输入一个精度,程序会计算出正奇数的倒数的正负交替数列和的4倍结果。 [2024-10-30 17:40:09 | AI写代码神器 | 297点数解答]