u Calculate e
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
················计算出前9项,按照格式输出就可以。代码:
#include<cstdio> int main(){ int f[11]; f[0]=1; for(int i=1;i<11;i++) f[i]=f[i-1]*i; printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");//前2项直接复制粘贴······ double x=0; for(int i=0;i<3;i++) x+=(double)1/f[i]; for(int i=3;i<10;i++){ x+=(double)1/f[i]; printf("%d %.9f\n",i,x); } return 0; }
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时间: 2024-11-05 02:38:20