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u Calculate e
u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32719 Accepted Submission(s): 14683
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源代码一：#include <stdio.h>#include <stdlib.h>double fac(int n){ double c; if(n==1 || n==0) { c=1; return(c); } else { c=fac(n-1)*n;return(c); }}int main(){ double sum,t; int i,n,j=3; printf("n e\n"); printf("- -----------\n"); printf("0 1\n"); printf("1 2\n"); printf("2 2.5\n"); while(j<10) { for(i=0,sum=0;i<=j;i++) { t=fac(i); sum=sum+1/t; } printf("%d %11.9f\n",j,sum); j++; } system("pause"); return 0;}源代码二：#include <stdio.h>#include <stdlib.h>int main(){ double arr[10] = {1},result; int i = 1,j = 3; while(i < 10) { arr[i]=i*arr[i-1]; i++; } printf("n e\n"); printf("- -----------\n"); printf("0 1\n"); printf("1 2\n"); printf("2 2.5\n"); result = 2.5; while(j < 10) { result = result + 1/arr[j]; printf("%d %11.9f\n",j,result); j++; } system("pause"); return 0;}
u Calculate e
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