LightOJ 1245 Harmonic Number (II)

最新推荐文章于 2020-09-21 17:32:50 发布

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本文介绍了一种求解'1234-HarmonicNumber'问题的方法，通过计算n/i的前n项和来找到H(n)。文章提供了一个有效的解决方案，并给出了具体的样例输入输出，帮助理解实现细节。

题目链接:点我

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 < 2^31).

Output

For each case, print the case number and H(n) calculated by the code.

Sample Input

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:

如代码所示,求n / i 的前n项和.

思路:

对于本题可以将它转化为，在坐标轴中y=n/x和x=y两条线形成的图形中，类似梯形中的整数点的个数，即为我们要求的，在最后需要减去边长为(int )sqrt(n)正方形的面积,因为它被计算了两次.

代码:

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

int main(){
    int t;
    scanf("%d", &t);
    int kase = 0;
    while(t--){
        int n;
        scanf("%d", &n);
        long long ans = 0;
        int i;
        for(i = 1; i <=(int)sqrt(n); ++i)
            ans += n/i;
        ans *= 2;
        ans -=(i - 1)*(i - 1);
        printf("Case %d: %lld\n",++kase, ans);
    }
    return 0;
}