HDOJ 1012 u Calculate e 解题报告

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23177    Accepted Submission(s): 10216

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

 

 

这道题目也没什么，

初看觉得有思路就开始写，

写着写着写到最后发现了输出上有点问题，，

不过还是解决了，

这个牛人就多了，，，

有个代码直接输出，霸气

#include<cstdio>
#include<cstdlib>

int main() 
{
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    printf("3 2.666666667\n");
    printf("4 2.708333333\n");
    printf("5 2.716666667\n");
    printf("6 2.718055556\n");
    printf("7 2.718253968\n");
    printf("8 2.718278770\n");
    printf("9 2.718281526\n");
    return 0;
}

 

我自己的代码就显得长了很多

#include<stdio.h>
float func(int n);
int main()
{
    int i, j, a[10];
    float sum=0;
    for(i=0;i<10;i++)
        a[i]=i;
    printf("n e\n- -----------\n");
    for(i=0;i<10;i++)
    {
        for(j=0;j<=i;j++)
        {
            sum+=1/func(a[j]);
        }
    if(i==0 || i==1)
        printf("%d %.0f\n", i, sum);
    else if(i==2)
        printf("%d %.1f\n", i, sum);
    else
        printf("%d %.9f\n", i, sum);
    sum=0;
    }
}
float func(int n)
{
    if (n==0)
        return 1;
    else 
        return n*func(n-1);
}

这个也是一个办法，

把前面几个直接输出，

再考虑后面的情况

#include<stdio.h>
int main()
{
    int i,j,k;
    double sum;
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n1 2\n2 2.5\n");
    sum=2.5;
        for(i=3;i<=9;i++)
        {
            k=1;
            for(j=1;j<=i;j++)
           {
                  k*=j;
           }
           sum+=1.0/k;
          printf("%d %.9lf\n",i,sum);
        }
    return 0;
}