【调和级数 && 欧拉常数】 Harmonic Number

• Step1 Problem:

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

• Step2 Ideas:

 ^{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}

也可以用打表水过去。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。

• Step3 code：

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define lt k<<1
#define rt k<<1|1
#define lowbit(x) x&(-x)
#define lson l,mid,lt
#define rson mid+1,r,rt
using namespace std;
typedef long long  ll;
typedef long double ld;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define mem(a, b) memset(a, b, sizeof(a))
//#define int ll
const double pi = acos(-1.0);
const double eps = 1e-6;
const double C = 0.57721566490153286060651209;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 1e5 + 5;
double a[maxn];

int main()
{
    a[1] = 1.0;
    for (int i = 2; i < 1e5; i++) a[i] = a[i - 1] + 1.0 / i;
    int T;
    cin >> T;
    for (int t = 1; t <= T; t++) {
        cout << "Case " << t << ": ";
        int n;
        cin >> n;
        if (n < 1e5) cout << fixed << setprecision(10) << a[n] << endl;
        else {
            double x = log(n) + C + 1.0 / (2.0 * n);
            cout << fixed << setprecision(10) << x << endl;
        }
    }
    return 0;
}

• Step4 code2：参考博客

#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
", ca++, s);
    }
    return 0;
}