--Number theory----to be complicated by a problem

Calculate N/1 + N/2 + N/3 +n/4 + ... + n/n =?

The first idea is to extract N:

Primitive = N (1+ 1/2 + 1/3 + 1/4 + ... 1/n) = n * (ln (n) + c) C is Euler constant, approx. 0.5772, sample hangs

And then found that this will be added to those that are not divisible by the omitted score, such as 2/3 = 0, but in the above formula is counted, go to the bug again, find extra number on the line

Not realized ...

From the Internet a look, far less difficult!

And then he knocked himself over:

#include <stdio.h> #include <string.h> #include <algorithm>using namespace Std;int main () {    long Long T,n;    scanf ("%lld", &t);    while (t--) {        scanf ("%lld", &n);        Long long  s = 0;        Long long ID;        For (long long i = 1;i<= n;) {             id = n/i;             s + = ID;             if (n/id = = id) i++;             else{                s+= (n/id-i) *id;                i = n/id + 1;             }        }        printf ("%lld\n", s);    }}

　　

--Number theory----to be complicated by a problem