LightOJ 1245 Harmonic Number (II)

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70017#problem/G

Harmonic Number (II)

Description

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：给定一个n，让你求Σn/i，i从1->n.

解析：例如10

　　　那么10/1=10;

　　　　　10/2=5;

　　　则n/i为1的数和为 1*（10-5）;　　

　　　同时对应着n/1的数为36，是以两段对应和为5+1*（10-5）;

　　　同理，10/3=3， 和为 5+2*（5-3）

　　　　...

　　　当n/i和i产生重合或者交叉时，就可以退出策画了

　　　全部策画流程如下

　　　i      1   2   3   4   5   6   7   8   9   10

　　　n/i  10  5   3   2   2   1   1   1   1    1

       如上所示，在策画i=1时，ans+=10， ans+=10-5

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <cstdlib>#include <limits>#include <queue>#include <stack>#include <vector>#include <map>using namespace std;#define N 10005000#define INF 0x3f3f3f3f#define PI acos (-1.0)#define EPS 1e-8#define P (1000000000+7)#define met(a, b) memset (a, b, sizeof (a))typedef long long LL;int main (){    int t, n, nCase = 1;    scanf ("%d", &t);    while (t--)    {        scanf ("%d", &n);        LL ans = 0, i;        for (i=1; i<=sqrt (n); i++)        {            ans += n / i;            if (n / i > n / (i+1))                ans += (n / i - n / (i+1)) * i;        }        if (n / (i - 1) == i - 1)//判断重合的时候要减去重复的            ans -= n / (i - 1);        printf ("Case %d: %lld\n", nCase++, ans);    }    return 0;}