输入一个正整数n，计算1+1/(1+2)+1/(1+2+3)+·· · ·+1/(1+2+3+··· · ·+n)的值(2 < n < 1000, 结果保留两位小数)。c++

以下是C++代码示例，用于计算1+1/(1+2)+1/(1+2+3)+···+1/(1+2+3+···+n)的值：

#include <iostream>
using namespace std;

int main() {
    int n;
    cout << "请输入一个正整数：";
    cin >> n;

    float sum = 0;
    float total = 0;
    for (int i = 1; i <= n; i++) {
        total += i;
        sum += 1 / total;
    }

    cout << "结果为：" << fixed << setprecision(2) << sum << endl;

    return 0;
}