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1234 - Harmonic Number

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Time Limit: 3 second(s)Memory Limit: 32 MB

In mathematics, the n^th harmonic number isthe sum of the reciprocals of the firstn 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 than10^-8 will be ignored.

Sample Input

Output for Sample Input

12

1

2

3

4

5

6

7

8

9

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;}