LightOJ 1245 Harmonic Number (II)（找规律）

http://lightoj.com/volume_showproblem.php?problem=1245

G - Harmonic Number (II)

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1245

Description

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

根据题中的代码便可知道题意，题意不多说

 

先看两个例子

1.

n = 10    sqrt(10) = 3     10/sqrt(10) = 3

i        1   2   3         4   5   6   7   8   9   10

n/i    10  5   3         2   2   1   1   1   1    1

 

m =  n/i

sum += m;

m = 1的个数10/1-10/2 = 5；

m = 2的个数10/2-10/3 = 2；

m = 3的个数10/3-10/4 = 1；

 

2.

n = 20     sqrt(20) = 4     20/sqrt(20) = 5

i        1   2   3   4       5   6   7   8   9   10   11   12   13   14   15   16   17   18   19   20

n/i    20  10 6   5       4   3   2   2   2    2     1     1     1     1     1     1     1     1    1    1

 

m =  n/i

sum += m;

m = 1的个数20/1-20/2 = 10；

m = 2的个数20/2-20/3 = 4；

m = 3的个数20/3-20/4 = 1；

m = 4的个数20/4-20/5 = 1；

...

m = i的个数20/i - 20/(i + 1)(1<= i <= sqrt(n))

 

这样我们可以得出：sqrt(n)之前的数我们可以直接用for循环来求

sqrt(n)之后的sum += (n/i - n/(i + 1)) * i;

当sqrt(n) = n / sqrt(n)时(如第一个例子10，sum就多加了一个3)，sum多加了一个sqrt(n),减去即可;

 

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>

using namespace std;
const int N = 10010;
typedef long long ll;

int main()
{
    int t, n, p = 0;
    ll sum;
    scanf("%d", &t);
    while(t--)
    {
        sum = 0;
        p++;
        scanf("%d", &n);
        int m = sqrt(n);
        for(int i = 1 ; i <= m ; i++)
            sum += n / i;
        for(int i = 1 ; i <= m; i++)
             sum += (n / i - n / (i + 1)) * i;
        if(m == n / m)
            sum -= m;
        printf("Case %d: %lld
", p, sum);
    }
    return 0;
}