【数论】 HDOJ 4986 Little Pony and Alohomora Part I

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 40005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
//typedef int LL;
using namespace std;
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

double res[maxn];
const double e = 2.718281828459045;
int main(void)
{
int n;
double ans;
for(int i = 1; i < maxn; i++) res[i] = res[i-1] + 1.0/i;
while(scanf("%d", &n)!=EOF) {
if(n < 100000) printf("%.4f\n", res
);
else {
ans = log((double)n) + 0.5772115;
printf("%.4f\n", ans);
}
}
return 0;
}