LightOJ 1245 Harmonic Number (II)(找规律)

时间:2021-07-09 14:29:28

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

G - Harmonic Number (II)

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

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