首页 > 代码库 > poj 1517 & hdu 1012 u Calculate e（简单阶乘）
poj 1517 & hdu 1012 u Calculate e（简单阶乘）
POJ链接 ：http://poj.org/problem?id=1517
HDU链接：http://acm.hdu.edu.cn/showproblem.php?pid=1012
Description
A simple mathematical formula for e is
e=Σ0<=i<=n1/i!
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.
Input
No input
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 Input
no input
Sample Output
n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333 ...
Source
Greater New York 2000
注意：poj上用G++交WA了，用C++AC，HDU上则不存在此问题。
代码如下：
#include <cstdio>
int Fac(int n)
{
int s=1;
for(int i=1; i<=n; i++)
s*=i;
return s;
}
int main()
{
double sum;
printf("n e\n- -----------\n");
for(int i = 0; i <= 9; i++)
{
sum = 1;
for(int j = 1; j <= i; j++)
{
sum+=1.0/Fac(j);
}
if(i <= 1)
printf("%d %.0lf\n",i,sum);
else if(i == 2)
printf("2 2.5\n");
else if(i > 2)
printf("%d %.9lf\n",i,sum);
}
return 0;
}
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