Harmonic Number(调和级数+欧拉常数)

Harmonic Number

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

Description

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

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

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

^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: My New Blog}

^{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 <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;
}

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;
}