Harmonic Number (II)

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： 计算n/1+n/2+…+n/n的值

思路： 题所给数据范围太大，故用题中所给方法暴力肯定是不行的，所以要找规律，我看其他大佬的博客学会了两种

1.大佬博客

y=n/x，是一个反比例函数，关于y=x对称，所以我们求n/x的和，相当于求图形的面积，可以分成两份一样的面积（在y=x处分，即一份x从1到sqrt(n),一份从sqrt(n)到n），此时x=sqrt(n),但两份面积有重叠，所以要减去重叠部分面积sqrt(n)*sqrt(n)（可能光看这些文字想不明白，所以可去看上述博客里的图片，一目了然）
2.大佬博客
这第二种方法就是找sqrt(n)到n之间n/x的和，前半段可暴力，不会超时，后半段按规律相加即可，上述博客叙述的很详细，我就不再多赘述了

AC代码：

第一种：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	int kk=1;
	while(t--)
	{
		int x;
		scanf("%d",&x);
		long long sum=0;
		int m=sqrt(x);
		for(int i=1; i<=m; i++)//前半段面积
			sum+=x/i;
		sum=sum*2-m*m;//*2-重叠面积
		printf("Case %d: %lld\n",kk++,sum);
	}

第二种：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	int t,n,i,j=1,k;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		k=sqrt(n);
		long long sum=0;
		for(i=1;i<=k;i++)//前半段和
			sum+=n/i;
		for(i=1;i<=k;i++)//后半段和
		sum+=(n/i-n/(i+1))*i;
		if(n/k==k)//减算重的
		sum-=k;
		printf("Case %d: %lld\n",j++,sum);
	}
	return 0;
}