Harmonic Number(調和級數+歐拉常數)

題意：求f(n)=1/1+1/2+1/3+1/4…1/n   (1 ≤ n ≤ 10⁸).，精確到10^{-8    (原題在文末）}

^知識點：

     ^{調和級數(即f(n))至今沒有一個完全正確的公式，但歐拉給出過一個近似公式：(n很大時)}

      f(n)≈ln(n)+C+1/2*n    

      ^{歐拉常數值：C≈0.57721566490153286060651209}

      ^{c++ math庫中，log即為ln。}

 

^題解:

^{公式：f(n)=ln(n)+C+1/(2*n);}

^{n很小時直接求，此時公式不是很准。}

 

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double r=0.57721566490153286060651209;     //歐拉常數
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;
}

 

其實可以打表水過,畢竟公式記不住是硬傷啊。。

10e8全打表必定MLE，而每40個數記錄一個結果，即分別記錄1/40,1/80,1/120,...,1/10e8，這樣對於輸入的每個n，最多只需執行39次運算，大大節省了時間，空間上也夠了。

#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