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
