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题目链接：

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

 

题目翻译：

求n/1+n/2+n/3+n/4+......+n/n的和。

 

规律题，需要自己手推一下公式，有点难想。（题解来自网上，比较好懂）

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 再求出值为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。  

样例分析：

n = 10，sqrt(10) = 3，10/sqrt(10) = 3i      1  2  3     4 5 6 7 8 9 10n/i    10 5  3     2 2 1 1 1 1 1m=n/isum += 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；

代码实现：

 

#include
   
    #include
    
     #define ll long longusing namespace std;int main(){	int T,kcase=1;	ll n,i;	scanf("%d",&T);	while(T--){		scanf("%lld",&n);		ll ans=0,m=(ll)sqrt(n);		for(i=1;i<=m;++i){			ans+=n/i;			ll l=n/i;			ll r=n/(i+1);			if(l>r)				ans+=(l-r)*i;		} 		if(n/m==m)//重复计算的减去 			ans-=m;		printf("Case %d: %lld\n",kcase++,ans);	}	return 0;}
    
   

 

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