HDU 1012 u Calculate e(數學題)

                                                                  u Calculate e

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

Total Submission(s): 40234    Accepted Submission(s): 18274

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
        

題解： 數學題。。。。輸出n=0~9時的結果。

AC代碼：

#include<iostream>
#include<cstdio>
int f(int n)
{
	int i,t;
	if(n==0) return 1;
	t = 1;
	for(i=1;i<=n;i++) t *= i;
	return t;
} 
int main()
{
	int i,j,n;
	double t,r;
	printf("n e\n");
	printf("- -----------\n");
	for(i=0;i<=9;i++)
	{
		t = 0.0;
		for(j=0;j<=i;j++)
			t += 1.0/f(j);
		if(i==0||i==1)printf("%d %.0f\n",i,t);
		else if(i==2)printf("%d %.1f\n",i,t);
		else printf("%d %.9lf\n",i,t);
	}
	return 0;
}
           

奇葩代碼：

#include<cstdio>
int main()
{
printf("n e\n");
printf("- -----------\n");
printf("0 1\n1 2\n2 2.5\n");
printf("3 2.666666667\n");
printf("4 2.708333333\n");
printf("5 2.716666667\n");
printf("6 2.718055556\n");
printf("7 2.718253968\n");
printf("8 2.718278770\n");
printf("9 2.718281526\n");

return 0;