HDU 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

 

 

Source

Greater New York 2000

 

 

 

看着有点吓人 但是很水 按照题目公式来就好了

 

/*
    前三个数 直接打印就行 
*/
#include<cstdio>
#include<iostream>

using namespace std;

int main() {
    printf("n e\n");
    printf("- ");
    for(int i=1;i<=11;i++) printf("-");
    printf("\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    for(int i=3;i<=9;i++) {
        double ans=.0,k=1;
        int t=0;
        printf("%d ",i);
        for(int j=0;j<=i;j++) {
            ans+=(1/k);
            t++;
            k*=t;
        }
        printf("%.9lf\n",ans);
    }
    return 0;
}