Harmonic Number (II)
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
//很好的题解：
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。
#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
int t,n,i,j;
int T=1;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
long long ans=0;
for(i=1;i<=(int)sqrt(n);i++)
{
ans+=n/i;
if(n/i>n/(i+1))
ans+=(long long)((n/i-n/(i+1))*i);
}
i--;
if(n/i==i)
ans-=i;
printf("Case %d: %lld\n",T++,ans);
}
return 0;
}
