Harmonic Number (II) 数学找规律

I was trying to solve problem '1234 - Harmonic Number', I wrote the following code

long long H( int n ) {
    long long res = 0;
    for( int i = 1; i <= n; i++ )
        res = res + n / i;
    return res;
}

Yes, my error was that I was using the integer divisions only. However, you are given n, you have to find H(n) as in my code.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n < 2³¹).

Output

For each case, print the case number and H(n) calculated by the code.

Sample Input

11

1

2

3

4

5

6

7

8

9

10

2147483647

Sample Output

Case 1: 1

Case 2: 3

Case 3: 5

Case 4: 8

Case 5: 10

Case 6: 14

Case 7: 16

Case 8: 20

Case 9: 23

Case 10: 27

Case 11: 46475828386

思路： 找规律这件事，emmm.....

　　　注意sqrt(n)这个数，数之间的差与后面数的个数。。。。。

　　　写几个完整的例子，努力寻找规律！

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 #include <math.h>
 #include <algorithm>
 #include <queue>
 #include <map>
 #include <stack>
 #include <deque>
 #include <iostream>
 using namespace std;
 typedef long long LL;
 const LL N = ;

 map<int, bool> check;
 int prime[];

 long long H( int n ) {
     long long res = ;
     for( int i = ; i <= n; i++ )
         res = res + n / i;
     return res;
 }

 int main()
 {
     LL i, p, j, n, t, cnt = ;
     LL sum;

     scanf("%lld", &t);
     while(t--) {
         sum = ;
         scanf("%lld", &n);
         for(i = ; i <= (LL)sqrt(n); i++) {
         //    cout << "i: " << i << endl;
             sum += (n / i - n / (i + )) * i;
             sum += n / i;
         }
         if(n / (LL)sqrt(n) == (LL)sqrt(n)) {
             // sum -= (n / (LL)(sqrt(n)) - n / (LL)(sqrt(n) + 1)) * (i - 1);
             sum -= (LL)sqrt(n);
         }
         printf("Case %lld: %lld\n", cnt ++, sum);
     }
     return ;

     //2 147 483 648
 }