HDOJ(递归)——1012.u Calculate e
Problem Description
A simple mathematical formula for e is
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.
Sample Output
n e
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333
题目大意:
给定e的计算公式,按格式输出n为0-9e的结果。
题目解析:
简单递归可以计算出e的值,要注意输出格式。
具体代码:
#include<iostream>
#include<algorithm>
using namespace std;
int getnum(int n){
if(n==0||n==1)
return 1;
else
return n*getnum(n-1);
}
int main()
{
printf("n e\n");
printf("- -----------\n");
printf("0 1\n");
printf("1 2\n");
printf("2 2.5\n");
printf("3 2.666666667\n");
printf("4 2.708333333\n");
double sum=0;
for(int i=0;i<=9;i++){
int fm=getnum(i);
sum+=(double)1/fm;
if(i>4)
printf("%d %.9f\n",i,sum);
}
return 0;
}