u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35178 Accepted Submission(s):
15843
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
//水题,格式;
1 #include<stdio.h> 2 double sieve[100]; 3 int main() 4 { 5 int i, total=1; sieve[0]=1; 6 for(i=1;i<10;i++) 7 { 8 total*=i; 9 sieve[i]=sieve[i-1]+1.0/total; 10 } 11 printf("n e\n"); 12 printf("- -----------\n"); 13 int n; 14 for(i=0;i<10;i++) 15 { 16 printf("%d ",i); 17 if(i<=1) 18 printf("%d\n",(int)sieve[i]); 19 else if(i==2) 20 printf("%.1lf\n",sieve[i]); 21 else 22 printf("%.9lf\n",sieve[i]); 23 } 24 return 0; 25 }