HDOJ 1012 u Calculate e
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来源:http://acm.hdu.edu.cn/showproblem.php?pid=1012
u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25914 Accepted Submission(s): 11477
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 11 22 2.53 2.6666666674 2.708333333
Source
Greater New York 2000
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题意:利用给出的公式计算e 的值。
题解:先把所有变量定义成double 型,对于n取0,1,2的情况就直接用cout输出,因为他们的小数位不一样,直接输出比较简单。后面的直接用公式计算便可,采用printf控制小数位输出。
PS:....
#include<iostream>using namespace std;int main(){double e=2.5,n,sum=1,i,j;cout<<"n e"<<endl; cout<<"- -----------"<<endl;cout<<"0"<<" "<<"1"<<endl;cout<<"1"<<" "<<"2"<<endl;cout<<"2"<<" "<<"2.5"<<endl; for(i=3;i<10;i++){for( j=1;j<=i;j++){sum*=j;}e+=1/sum;printf("%.0lf %.9lf\n",i,e);sum=1; }return 0;}
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