ＨＤＵ １０１２ u Calculate e 水题 基础题

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 11 22 2.53 2.6666666674 2.708333333

 

Source

Greater New York 2000

 

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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;}