Harmonic Number(调和级数+欧拉常数)

时间：2021-01-02 06:57:45

题意：求f(n)=1/1+1/2+1/3+1/4…1/n   (1 ≤ n ≤ 108).，精确到10-8    (原题在文末）

知识点：

     调和级数(即f(n))至今没有一个完全正确的公式，但欧拉给出过一个近似公式：(n很大时)

      f(n)≈ln(n)+C+1/2*n

      欧拉常数值：C≈0.57721566490153286060651209

      c++ math库中，log即为ln。

题解:

公式：f(n)=ln(n)+C+1/(2*n);

n很小时直接求，此时公式不是很准。

#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;

const double r=0.57721566490153286060651209;     //欧拉常数

double a[10000];

int main()

{

    a[1]=1;

    for (int i=2;i<10000;i++)

    {

        a[i]=a[i-1]+1.0/i;

    }

    int n;

    cin>>n;

    for (int kase=1;kase<=n;kase++)

    {

        int n;

        cin>>n;

        if (n<10000)

        {

            printf("Case %d: %.10lf\n",kase,a[n]);

        }

        else

        {

            double a=log(n)+r+1.0/(2*n);

            //double a=log(n+1)+r;

            printf("Case %d: %.10lf\n",kase,a);

        }

    }

    return 0;

}

其实可以打表水过,毕竟公式记不住是硬伤啊。。

10e8全打表必定MLE，而每40个数记录一个结果，即分别记录1/40,1/80,1/120,...,1/10e8，这样对于输入的每个n，最多只需执行39次运算，大大节省了时间，空间上也够了。

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <cmath>

using namespace std;

const int maxn = 2500001;

double a[maxn] = {0.0, 1.0};

int main()

{

    int t, n, ca = 1;

    double s = 1.0;

    for(int i = 2; i < 100000001; i++)

    {

        s += (1.0 / i);

        if(i % 40 == 0) a[i/40] = s;

    }

    scanf("%d", &t);

    while(t--)

    {

        scanf("%d", &n);

        int x = n / 40;

        s = a[x];

        for(int i = 40 * x + 1; i <= n; i++) s += (1.0 / i);

        printf("Case %d: %.10lf\n", ca++, s);

    }

    return 0;

}

Harmonic Number

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status

Description

In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:

Hn=1/1+1/2+1/3+1/4…1/n

In this problem, you are given n, you have to find Hn.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 108).

Output

For each case, print the case number and the nth harmonic number. Errors less than 10-8 will be ignored.

Sample Input

12

1

2

3

4

5

6

7

8

9

90000000

99999999

100000000

Sample Output

Case 1: 1

Case 2: 1.5

Case 3: 1.8333333333

Case 4: 2.0833333333

Case 5: 2.2833333333

Case 6: 2.450

Case 7: 2.5928571429

Case 8: 2.7178571429

Case 9: 2.8289682540

Case 10: 18.8925358988

Case 11: 18.9978964039

Case 12: 18.9978964139

相关文章