首页 > 代码库 > hdu 1012 u Calculate e
hdu 1012 u Calculate e
u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31968 Accepted Submission(s): 14314
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.
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: 1013 1017 1020 1014 1032
#include <stdio.h>
double fun ( int i )
{
double re = 2.5, to = 6;
int n;
for ( n = 3; n <= i; n++ )
{
re += 1 / to;
to *= ( n + 1 );
}
return re;
}
int main()
{
int i;
printf ( "n e\n" );
printf ( "- -----------\n" );
printf ( "0 1\n1 2\n2 2.5\n" );
for ( i = 3; i <= 9; i++ )
printf ( "%d %.9lf\n", i, fun ( i ) );
return 0;
}
hdu 1012 u Calculate e
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。