LightOJ 1245 （数论：找规律）

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=98634#problem/B

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：通过给出的代码计算i=1~n时n/i的和。

分析：举例n=8时，m=sqrt（n），得到所有的n/i为8，4，2，2，1，1，1，1，通过找规律可以发现有8-4个1，4-2个2，那么我们发现当i为m~n之间时是有规律的，我们可以当i在1~m时暴力，m~n时通过规律计算。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;

typedef long long LL;

LL Solve(LL n)
{
    int i, m, left, right;
    LL res = 0;

    m = (int)sqrt(n);

    for (i = 1; i <= m; i++)
        res += n/i; ///计算i=1~m的n/i和

    left = right = m; ///left，right是n/i==m的数的左右区间

    while (m > 0)
    {
        right = n/m; ///计算n/i==m的数的区间右端点
        res += (right-left)*m; ///计算该区间的和
        m--;
        left = right; ///更新左端点
    }

    return res;
}

int main ()
{
    int T, k = 0;
    LL n, ans;

    scanf("%d", &T);

    while (T--)
    {
        k++;

        scanf("%lld", &n);

        ans = Solve(n);

        printf("Case %d: %lld
", k, ans);
    }

    return 0;
}