C++ 程序：计算 1/1 - 1/2 + 1/3 - 1/4 + ... + (-1)^(n-1) * 1/n 的值

#include "iostream"\nusing namespace std;\n\nint main() {\n int n;\n cin >> n;\n \n double sum = 0.0;\n double sign = 1.0;\n for (int i = 1; i <= n; i++) {\n sum += sign * (1.0 / i);\n sign *= -1;\n }\n \n cout << fixed << setprecision(4) << sum << endl;\n \n return 0;\n}