POJ 1517 u Calculate e
u Calculate e
Time Limit: 1000MS | Memory Limit: 10000K | |||
Total Submissions: 16932 | Accepted: 10033 | Special Judge |
Description
A simple mathematical formula for e ise=Σ0<=i<=n1/i! where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
Input
No input
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 Input
no input
Sample Output
n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333 ...
这根本就是a+b的变形题么....
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 3 int main() 4 { 5 printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n3 2.666666667\n4 2.708333333\n5 2.716666667\n6 2.718055556\n7 2.718253968\n8 2.718278770\n9 2.718281526\n"); 6 return 0; 7 }