HDOJ-1012(u Calculate e)(输出格式)

HDOJ-1012(u Calculate e)(输出格式)

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37915    Accepted Submission(s): 17190

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

 

注：这道题不难，但是有几个坑。

/*这道题是坑,输出的实数的小数位长度为9,而2.5的后缀0,不输出,而i=8时,有一个后缀0却输出*/ 
/*刚开始以为小数点后的后缀0都不输出,结果提交总是wa,看了别人的博客才知道,i=8时的后缀0不输出*/ 

My  solution:

/*2015.11.15*/

#include<stdio.h>int njie(int b){    int t=1,i;     for(i=1;i<=b;i++)    t*=i;     return t;}int main(){        int n,i,j;    double sum=2,path;    printf("n e\n");    printf("- -----------\n");    printf("0 1\n");    printf("1 2\n");    for(i=2;i<=9;i++)    {        sum+=1.0/njie(i);        if(sum==2.5)        printf("%d 2.5\n",i);        else  //if(i==8)       // printf("%d %.8lf\n",i,sum);        //else        printf("%d %.9lf\n",i,sum);    }        return 0;}