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 = 10000010;

map<int, bool> check; 
int prime[7000000];

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

int main()
{
    LL i, p, j, n, t, cnt = 1;
    LL sum;

    scanf("%lld", &t);
    while(t--) {
        sum = 0;
        scanf("%lld", &n);
        for(i = 1; i <= (LL)sqrt(n); i++) {
        //    cout << "i: " << i << endl;
            sum += (n / i - n / (i + 1)) * 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
", cnt ++, sum);
    }
    return 0;

    //2 147 483 648
}