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

题目大意：用递归的方法去求H（n）太费事了，请找到规律，减少求H（n）的时间

解体思路：以100为例，通过列举几项发现，求H（n），即求有多少个1，多少个2，多少个3，一直到n/2，最后加上n。

设有x个i，则x*i=(n/i-n/(i+1))*i

这将一个大的数折成一半了（循环长度为n/2），然而还是会超时。因此继续优化，

我们可以将循环长度减到sqrt（n);从1-100，100-1同时计算，但会出现重复计算的地方，只要剪去就可以了。

代码如下：

#include<stdio.h>
#include<cmath>
int main()
{
long long r,t,i,s,sum,j;
scanf("%lld",&t);
j=1;
while(t--)
{
scanf("%lld",&r);
s=sqrt(r);
sum=0;
for(i=1;i<s+1;i++)
{
sum+=(r/i-r/(i+1))*i+r/i;
if(r/i==i)//出现重叠的地方
sum-=i;
}
printf("Case %lld: %lld\n",j++,sum);
}
return 0;
}