hdu 1012:u Calculate e（数学题，水题）

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28686 Accepted Submission(s): 12762


[align=left]Problem Description[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Output[/align]

n e

- -----------

0 1

1 2

2 2.5

3 2.666666667

4 2.708333333

[align=left]Source[/align]
Greater New York 2000

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　　水题。
　　切一道水题，放松下心情。
　　这道题没有输入，只有输出。
　　前3组数据由于输出格式不统一，直接输出即可。后面的数可用迭代思路求得，不用从头重新计算了。
　　代码：

#include <stdio.h>
int main()
{
int i;
printf("n e\n");
printf("- -----------\n");
printf("0 1\n");    //没有统一格式，提前输出。
printf("1 2\n");
printf("2 2.5\n");
double ans = 1;
double t = 1;
for(i=1;i<=9;i++){
t = 1.0/i*t;    //迭代计算
ans += t;
if(i<3) continue;
printf("%d %.9lf\n",i,ans);
}
return 0;
}

Freecode : www.cnblogs.com/yym2013