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

Source

Problem Setter: Jane Alam Jan

/很好的题解：转载的

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
    
     #include
     
      #include
      
        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;}