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

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

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

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；

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 ;

}

秒客网

LightOJ 1245 Harmonic Number (II)（找规律）

相关文章