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
![](http://acm.hdu.edu.cn/data/images/1012-1.gif)
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
![](http://acm.hdu.edu.cn/data/images/1012-1.gif)
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
#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; }