joj1020

1020: u Calculate e


ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE
3s 8192K 6098 1521 Standard

Background


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<stdio.h>
double factoral(double);
double showsum(int );

int main(void)
{
int n, i;
double sum = 0;

printf("n e\n");
printf("- -----------\n");

printf("0 1\n");
printf("1 2\n");
printf("2 2.5\n");


for (i=3; i<=9; i++)
{
printf("%d ", i);
printf("%.9lf\n",1+showsum(i));
}

return 0;
}

double factoral(double i)
{
if (i == 0)
return 1;
else
return i*factoral(i-1);
}

double showsum(int n)
{
double sum=0;
for(int i=1;i<=n;i++)
{
sum+=(double)1/factoral(i);
}
return sum;
}


 


This problem is used for contest: 76 182


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posted @ 2011-11-02 18:14  漂木  阅读(222)  评论(0编辑  收藏  举报