題意：求f(n)=1/1+1/2+1/3+1/4…1/n (1 ≤ n ≤ 108).，精確到10-8 (原題在文末） 知識(shí)點(diǎn)：調(diào)和級(jí)數(shù)(即f(n))至今沒(méi)有一個(gè)完全正確的公式，但歐拉給出過(guò)一個(gè)近似公式：(n很大時(shí))

f(n)≈ln(n)+C+1/2*n

歐拉常數(shù)值：C≈0.57721566490153286060651209

c++ math庫(kù)中，log即為ln。

題解:公式：f(n)=ln(n)+C+1/(2*n); n很小時(shí)直接求，此時(shí)公式不是很準(zhǔn)。 或者用打表做

題目：

In mathematics, the?nth?harmonic number is the sum of the reciprocals of the first?n?natural numbers:

In this problem, you are given?n, you have to find?Hn.

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?nth?harmonic number. Errors less than?10-8?will be ignored.

Sample Input

12

1

2

3

4

5

6

7

8

9

90000000

99999999

100000000

Sample Output

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

#include <iostream> #include <cstdio> #include <cmath> using namespace std; const double r=0.57721566490153286060651209; //歐拉常數(shù) double a[10000]; int main() {a[1]=1;for (int i=2;i<10000;i++){a[i]=a[i-1]+1.0/i;}int n;cin>>n;for (int kase=1;kase<=n;kase++){int n;cin>>n;if (n<10000){printf("Case %d: %.10lf\n",kase,a[n]);}else{double a=log(n)+r+1.0/(2*n);printf("Case %d: %.10lf\n",kase,a);}}return 0; }

記不住公式，就打表做

#include<iostream> #include<stdio.h> #include<string.h> using namespace std; const int M=1e8; int t; int m; double w[1000010]={0.0}; int main() {cin>>t;int k=1;double s=0.0;for(int i=1; i<=M; i++){s+=1.0/i;if(i%100==0)w[i/100]=s;}while(t--){cin>>m;int a=m/100;double ans=w[a];for(int i=a*100+1; i<=m; i++)ans+=1.0/i;printf("Case %d: %.10lf\n",k++,ans);}return 0; }

?

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