用c++计算1-1/2+1/3-1/4+1/5+…+(-1)ⁿ⁻¹1/n

以下是用C++计算1-1/2+1/3-1/4+1/5+…+(-1)ⁿ⁻¹1/n的代码：

#include <iostream>
using namespace std;

int main() {
    int n;
    double sum = 0;
    cout << "请输入n的值：";
    cin >> n;
    for (int i = 1; i <= n; i++) {
        if (i % 2 == 1) {
            sum += 1.0 / i;
        } else {
            sum -= 1.0 / i;
        }
    }
    cout << "结果为：" << sum << endl;
    return 0;
}