Harmonic Number (II)

Harmonic Number (II)

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Time Limit: 3 second(s)

Memory Limit: 32 MB

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：就是计算所给的函数的值，但是咱们如果按函数那种算法会TLE，so

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

int main()
{
    int t, n, l = 1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        long long k = sqrt(n);
        long long res = 0;
        for(int i = 1; i < k+1; i++)
        {
            res += n /i;
            if(n/i > n / (i+1))
                res += (n/i - n / (i+1))*i;
            if((k*k + i - 1) == n)  // 在k*k到k*k+k-1中的数多加了k，so减去
            {
                res -= k;
            }
        }
        printf("Case %d: %lld
", l++, res);
    }
    return 0;
}