LightOJ 1245: Harmonic Number (II) （数学）

http://lightoj.com/volume_showproblem.php?problem=1245

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	Output for Sample Input
11 1 2 3 4 5 6 7 8 9 10 2147483647	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

题意分析：

求

解题思路：

题目范围较大，直接求或者打表都会超时，

可以拿24这个数来看看， 三个数分别为n ， i， n/i。

前两个数为24和12， 末尾有24-12个1，

同理末尾有12-8个2，

有8-6个3，

有6-4个4.

这样只要计算到，再考虑一下重复的情况，就能算出答案。

#include <stdio.h>
#include <math.h>
int main()
{
	int t=0, T;
	long long n, i, sum, temp, m;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%lld", &n);
		sum=0;
		temp=n;
		m=sqrt(n);
		for(i=1; i<=m; i++)
			sum+=n/i + (n/i - n/(i+1)) * i;
		i--;	
		if(n/i==m)
			sum-=m;
		printf("Case %d: %lld
", ++t, sum);
	}
	return 0;
}