lightoj1234 分段打表

最新推荐文章于 2021-08-09 14:33:00 发布

liyunlong41

最新推荐文章于 2021-08-09 14:33:00 发布

阅读量815

点赞数

CC 4.0 BY-SA版权

分类专栏：数论相关文章标签：数论

本文链接：https://blog.youkuaiyun.com/liyunlong41/article/details/49931567

数论相关专栏收录该内容

23 篇文章

订阅专栏

本博客介绍如何计算给定自然数n的第n个调和数，并提供了高效的求解算法，通过预处理和线性查找的方式实现，适用于大量数据处理场景。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1234 - Harmonic Number

PDF (English)

Statistics

Forum

Time Limit: 3 second(s)

Memory Limit: 32 MB

In mathematics, the n^th harmonic number isthe 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^thharmonic number. Errors less than 10^-8 will be ignored.

Sample Input

Output for Sample Input

90000000

99999999

100000000

Case 1: 1

Case 2: 1.5

Case 3: 1.8333333333

Case 4: 2.0833333333

Case 5: 2.2833333333

Case 6: 2.450

Case 7: 2.5928571429

Case 8: 2.7178571429

Case 9: 2.8289682540

Case 10: 18.8925358988

Case 11: 18.9978964039

Case 12: 18.9978964139

题意：不说了。

分析：一亿的数据量，可以每50个或者100个记录下来。

#include<bitset>
#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

inline int in()
{
    int res=0;char c;
    while((c=getchar())<'0' || c>'9');
    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
    return res;
}
const int N=1e8;
double a[N/50+50];
void init()
{
    double t=1.0;
    for(int i=2;i<=N;i++)
    {
        t += 1.0/i;
        if(i%50 == 0)
        {
            a[i/50]=t;
        }
    }
}
int main()
{
    int T=in(),ii=1;
    init();
    while(T--)
    {
        int n=in();
        double ans = a[n/50];
        for(int i=n/50*50+1;i<=n;i++)
        {
            ans += 1.0/i;
        }
        printf("Case %d: %.10lf\n",ii++,ans);
    }
    return 0;
}