Harmonic Number (II)

Harmonic Number (II)

LightOJ - 1245

名称	数量/来源
Time limit	3000 ms
Memory	32768 kB
Source	Problem Setter: Jane Alam Jan

题目描述：

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.

输入：

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 < 2^31).

输出：

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

样例输入：

11

1

2

3

4

5

6

7

8

9

10

2147483647

样例输出：

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

题目大意：

先求出前sqrt(n)项和：即n/1+n/2+…+n/sqrt(n)

再求出后面所以项之和.后面每一项的值小于sqrt(n),计算值为1到sqrt(n)的项的个数，乘以其项值即可快速得到答案

例如:10/1+10/2+10/3+…+10/10

sqrt(10) = 3

先求出其前三项的和为10/1+10/2+10/3

在求出值为1的项的个数为(10/1-10/2)个，分别是(10/10,10/9,10/8,10/7,10/6),值为2个项的个数（10/2-10/3）分别是（10/5,10/4）,在求出值为3即sqrt(10)的项的个数.

显然，值为sqrt(10)的项计算了2次,减去一次即可得到答案。当n/(int)sqrt(n) == (int)sqrt(n)时，值为sqrt(n)的值会被计算2次。

code:

#include<stdio.h>
#include<math.h>
int main()
{
	int ttt;
	scanf("%d",&ttt);
	for(int kk=1;kk<=ttt;kk++)
	{
		long long n;
		scanf("%lld",&n);
		int m=sqrt(n);
		long long ans=0;
		for(int i=1;i<=m;i++)
		{
			ans+=n/i;
		}
		for(int i=1;i<=m;i++)
		{
			ans+=(n/i-n/(i+1))*i;
		}
		if(m==n/m)ans-=m;
		printf("Case %d: %lld\n",kk,ans);
	}
	return 0;
 }