HDOJ 1012 u Calculate e

来源：http://acm.hdu.edu.cn/showproblem.php?pid=1012

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25914    Accepted Submission(s): 11477


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

 

Recommend

JGShining   |   We have carefully selected several similar problems for you:  1005 1004 1013 1021 1017 

                                                                                                                                                                                                                                                      题意：利用给出的公式计算e 的值。

题解：先把所有变量定义成double 型，对于n取0，1，2的情况就直接用cout输出，因为他们的小数位不一样，直接输出比较简单。后面的直接用公式计算便可，采用printf控制小数位输出。
PS：....

#include<iostream>
using namespace std;
int main()
{
double e=2.5,n,sum=1,i,j;
cout<<"n e"<<endl;  cout<<"- -----------"<<endl;
cout<<"0"<<" "<<"1"<<endl;
cout<<"1"<<" "<<"2"<<endl;
cout<<"2"<<" "<<"2.5"<<endl;
for(i=3;i<10;i++){
for( j=1;j<=i;j++){
sum*=j;
}
e+=1/sum;
printf("%.0lf %.9lf\n",i,e);
sum=1;
}
return 0;
}