HDU 1012 u Calculate e (水题)

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
 


#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
	double ans,sum;
	printf("n e\n");
	printf("- -----------\n");
	printf("0 1\n");
	ans=sum=1;
	for(int i=1;i<=9;i++) {
		ans*=i;
		sum+=1.0/ans;
		if(i==1) printf("%d %0.0lf",i,sum);
		else if(i==2) printf("%d %0.1lf",i,sum);
		else printf("%d %0.9lf",i,sum);
		printf("\n");
	}
	return 0;
}


你可能感兴趣的:(HDU 1012 u Calculate e (水题))