HDOJ 1012 u Calculate e（水题）

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35564    Accepted Submission(s): 16038

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

 

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#include<stdio.h>
int main(){
	int n;
	double e,sum,ssum;
	printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");
	sum=2.5; ssum=2;
	for(n=3;n<10;n++){
		ssum*=n; //阶乘
		sum+=1/ssum; //求和
		printf("%d %.9lf\n",n,sum);
	}
	return 0;
}

然后我在讨论区看见这位仁兄的，震慑住我了：

# include<stdio.h>
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;
}