u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46276 Accepted Submission(s): 21237
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
#include<stdio.h>int jc(int n) { int i,j=1; for(i=1;i<=n;i++) j=j*i; return j; } int main() { int n,i; double e=2.5; printf("n e "); printf("- ----------- "); printf("0 1 "); printf("1 2 "); printf("2 2.5 "); for(i=3;i<=9;i++)//为什么从3开始,是因为从3之后,小数点后有9位小数 { e=e+1.0/jc(i); printf("%d %.9lf ",i,e); } return 0; }