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HDU 1012

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1645    Accepted Submission(s): 696


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
#include<iostream>
using namespace std;

int main() 
{
    int f[10]={1,1,2},i;
    double e[10]={1.0,2.0,2.5};
    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){
        f[i]=f[i-1]*i;
        e[i]=e[i-1]+1.0/f[i];
        printf("%d %.9lf\n",i,e[i]);
    }
    return 0;
}

此题无甚，注意输出格式。
posted on 2007-11-29 08:15 tctony 阅读(600) 评论(0)  编辑 收藏 引用

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