u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35137 Accepted Submission(s): 15824
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
开始错了几次，后来在网上说第八个后面有个０，最后决定用ｐｒｉｎｔｆ控制格式
#include <iostream>
#include <iomanip>
#include <cstdio>
using namespace std;
double a;
double sum[10];
int main()
{
a=1;
sum[0]=1;
sum[1]=2;
for(int i=2; i<=9; i++)
{
a=a*(1.0/i);
sum[i]=a+sum[i-1];
}
cout<<"n e"<<endl;
cout<<"- -----------"<<endl;
cout<<"0 1"<<endl;
cout<<"1 2"<<endl;
cout<<"2 2.5"<<endl;
for(int i=3;i<=9;i++)
{
printf("%d %.9f
",i,sum[i]);
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。