34.设s=1+1/2+1/3+…+1/n,求与8最接近的s的值及与之对应的n值

#include<iostream>
using namespace std;

int main()
{
    int n=1;
    double s=0;
    do//先执行一次循环体
    {
        s=s+(1.0/n);
        n++;
    }while(s<8.0);
    cout <<"s="<<s<<endl;
    cout <<"n="<<n<<endl;

    return 0;

}

34.设s=1+1/2+1/3+…+1/n,求与8最接近的s的值及与之对应的n值