hdoj 1012 u Calculate e

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37350    Accepted Submission(s): 16905

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

 

按照上边的公式 从0求到9 输出即可                注：小技巧（我看到的想记录下来，与本题无关）printf输出的时候printf("%.5g",n);可以将小数点后边的没用的0去掉如 9.231000可输出9.231

 

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

#include<stdio.h> #include<string.h> double b[20]; double fun(int x) {     int i;     double sum=1.0,a=1.0;     for(i=1;i<=x;i++)     {         sum=a*sum;         a=a+1;     }     sum=1/sum;     return sum; } int main() {     int n,m,j,i;     double sum=0;     b[0]=1;b[1]=1;     for(i=2;i<=9;i++)         b[i]=fun(i);     printf("n e\n");     printf("- -----------\n");     for(i=0;i<=9;i++)     {         printf("%d ",i);         sum+=b[i];         if(i==2)         {             printf("2.5\n");             continue;         }         if(sum==(int)(sum))             printf("%d\n",(int)(sum));         else             printf("%.9lf\n",sum);     }     return 0; }