Harmonic Number(调和级数+欧拉常数)

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 n^th 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 H_n.

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 n^th 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