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Description

In mathematics, we suppose that f(1)=1, f(i)-f(i-1)=1/i, (2<=i<=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 (2 ≤ n ≤10^7).

Output

For each case, print the case number and the value of f(n). The answer should be rounded to 10 decimal places.

Sample Input

3234

Sample Output

Case 1: 1.5000000000Case 2: 1.8333333333Case 3: 2.0833333333 題意：求f(n)=1/1+1/2+1/3+1/4…1/n，精確小數(shù)點(diǎn)后10位

^{知識(shí)點(diǎn)：}

^{???? 調(diào)和級(jí)數(shù)(即f(n))至今沒有一個(gè)完全正確的公式，但歐拉給出過一個(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庫中，log即為ln。}

^{但是n很小是直接求}

^貼代碼：

 #include <bits/stdc++.h> using namespace std; typedef long long ll; #define inf 0x3f3f3f3f #define maxn 10000011 double r = 0.57721566490153286060651209; double a[10005]; int main() {a[1] = 1;for(int i=2; i<=10000; i++)a[i] = a[i-1] + 1.0/i;int t;scanf("%d",&t);for (int cas=1; cas<=t; cas++){int n;scanf("%d",&n);if(n <= 10000)printf("Case %d: %.10lf\n",cas,a[n]);else{double sum;sum = log(n) + r + 1.0/(2*n);printf("Case %d: %.10lf\n",cas,sum);}} }


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