u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53154 Accepted Submission(s): 24348
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
这道题目一开始的做法是计算出所有的数据存入double型的数组,然后用%.10g输出(%.10g表示输出结果在10位之内的有效数字,不包含无意义的0),但是由于当n为8时,结果是2.718278770,最后结果中含有无效的0,测试结果也要求输出,于是n等于3之后的结果用%lf输出。
#include<bits/stdc++.h>
using namespace std;
double del(int m)
{
double mul=1;
for(int i=1;i<=m;i++)
mul*=i;
return 1.0/mul;
}
int main()
{
double num[10];
num[0]=1;num[1]=2,num[2]=2.5;
for(int i=3;i<=9;i++)
{
num[i]=del(i)+num[i-1];
}
printf("n e\n- -----------\n");
printf("%d %g\n%d %g\n%d %g\n",0,num[0],1,num[1],2,num[2]);
for(int i=3;i<=9;i++)
printf("%d %.9lf\n",i,num[i]);
return 0;
}