u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28686 Accepted Submission(s): 12762
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.
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
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这道题没有输入，只有输出。
前3组数据由于输出格式不统一，直接输出即可。后面的数可用迭代思路求得，不用从头重新计算了。
代码：
1 #include <stdio.h>
2 int main()
3 {
4 int i;
5 printf("n e
");
6 printf("- -----------
");
7 printf("0 1
"); //没有统一格式，提前输出。
8 printf("1 2
");
9 printf("2 2.5
");
10 double ans = 1;
11 double t = 1;
12 for(i=1;i<=9;i++){
13 t = 1.0/i*t; //迭代计算
14 ans += t;
15 if(i<3) continue;
16 printf("%d %.9lf
",i,ans);
17 }
18 return 0;
19 }
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