ACM HDU 1012 u Calculate e

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19296 Accepted Submission(s): 8425

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

 

 

#include<stdio.h>
#include<string.h>
double run(int n)
{
    if(n == 0 || n == 1) return 1;
    else return run(n-1)*n;
}

int main()
{
    int n, i, j;
    double sum[11];
    memset(sum, 0, sizeof(sum));
    printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");
    sum[2] = 2.5;
    for(i=3; i<=9; i++)
    {
        sum[i] = sum[i-1] + 1.0/run(i);
        printf("%d %.9lf\n", i, sum[i]);
    }
}

结题报告：

WA的情况并没有那么唯一的，捉襟见肘的感觉很难见到AC的那一刻，做题需要严谨和耐心，WA多次没问题，挣扎成长只是暂时

如果没有这个提示，我想我会再禁锢一段时间

n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333
5 2.716666667
6 2.718055556
7 2.718253968
8 2.718278770
9 2.718281526

这是AC的数据，注意n=8的情况，最后面无意义的0还是要保留。所以n=0、1、2作为特殊值输出，剩下的用%.9lf即可，%.10lg就要WA了。