LightOJ 1245(Harmonic Number (II))

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

题意：求f(n)=n/1+n/2.....n/n,其中n/i保留整数

分析：看了一个大神的blog，f=n/x这个函数关于y = x 对称对称点刚好是sqrt(n)，

于是就简单了直接求sum+n/i (i*i<n && i >=1)，然后乘以2，再减去i*i即可。

#include<cstdio>
#include<cmath>
//f=N/x,对称点是sqrt(N) 
long long H(int N)
{
    int K=sqrt(N);
    long long ans=0;
    for(int i=1;i<=K;i++)
        ans+=N/i;
    ans*=2;
    ans-=K*K;
    return ans;
}

int main()
{
    int T,N,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        printf("Case %d: %lld
",cas++,H(N));
    }
    return 0;
}