u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 43467    Accepted Submission(s): 19919 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   Recommend JGShining   |   We have carefully selected several similar problems for you:   1004  1013  1021  1017  1020    基础题，水题，一开始在纠结牵前两个的输出格式怎么控制，想了半天，还i交了一遍，嗯，最后选择了直接输出，因为后面都是９位　　很好控制了 ａｃ代码 #include <iostream> #include <cstdio> using namespace std; int main() { double a[15]={1.0,2.0,2.5}; double jie[15]={1.0,1.0}; int i,j; for(i=2;i<=10;i++) jie[i]=jie[i-1]*i; for(i=2;i<=10;i++) { a[i]=a[i-1]+(1.0/jie[i]); } printf("n e\n"); printf("- -----------\n"); printf("0 1\n"); printf("1 2\n"); printf("2 2.5\n"); for(i=3;i<10;i++) { printf("%d %.9lf\n",i,a[i]); } return 0; }