LocalMaxima

it2022-05-07  0

先打表,发现\(ans=\sum_{i=1}^n\frac{1}{i}\)

对于小数据可以直接打表

数据很大时,精度相对就比较宽松

欧拉-马斯刻若尼常数=调和级数-自然对数

调和级数为:\(\sum_{i=1}^{\infty}\frac{1}{i}\) 自然对数就是:\(\ln (x)\)

欧拉-马斯刻若尼常数:\(\gamma=\lim _{n\to \infty}[(\sum_{i=1}^n)-\ln(n)]=\int_1^{\infty}(\frac{1}{\lfloor x\rfloor}-\frac {1}{x})dx\) 近似值约为:\(\gamma\approx 0.577215664901532860606512090082402431042159335\)

于是这道题就解决了

#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<cstring> #include<algorithm> #include<queue> #include<stack> #include<set> #include<bitset> #include<vector> #include<cstdlib> #include<ctime> #define QAQ int #define TAT long long #define OwO bool #define ORZ double #define F(i,j,n) for(QAQ i=j;i<=n;++i) #define E(i,j,n) for(QAQ i=j;i>=n;--i) #define MES(i,j) memset(i,j,sizeof(i)) #define MEC(i,j) memcpy(i,j,sizeof(j)) using namespace std; const ORZ pho=0.577215664901532; QAQ n; ORZ ans; QAQ main(){ scanf("%d",&n); if(n<=1000000) F(i,1,n) ans+=1.0/(ORZ)i; else ans=log(n)+pho; printf("%.8lf\n",ans); return 0; }

转载于:https://www.cnblogs.com/heower/p/8467816.html


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