题意：求f(n)=1/1+1/2+1/3+1/4…1/n (1 ≤ n ≤ 108).，精确到10-8 (原题在文末）
知识点：
调和级数(即f(n))至今没有一个完全正确的公式，但欧拉给出过一个近似公式：(n很大时)
f(n)≈ln(n)+C+1/2*n
欧拉常数值：C≈0.57721566490153286060651209
c++ math库中，log即为ln。
题解:
公式：f(n)=ln(n)+C+1/(2*n);
n很小时直接求，此时公式不是很准。
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double r=0.57721566490153286060651209; //欧拉常数
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);
//double a=log(n+1)+r;
printf("Case %d: %.10lf\n",kase,a);
}
}
return 0;
}
其实可以打表水过,毕竟公式记不住是硬伤啊。。
10e8全打表必定MLE，而每40个数记录一个结果，即分别记录1/40,1/80,1/120,...,1/10e8，这样对于输入的每个n，最多只需执行39次运算，大大节省了时间，空间上也够了。
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath> using namespace std; const int maxn = 2500001;
double a[maxn] = {0.0, 1.0}; int main()
{
int t, n, ca = 1;
double s = 1.0;
for(int i = 2; i < 100000001; i++)
{
s += (1.0 / i);
if(i % 40 == 0) a[i/40] = s;
}
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
int x = n / 40;
s = a[x];
for(int i = 40 * x + 1; i <= n; i++) s += (1.0 / i);
printf("Case %d: %.10lf\n", ca++, s);
}
return 0;
}
Harmonic Number
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Description
In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn=1/1+1/2+1/3+1/4…1/n
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