HDU 1012 u Calculate e(水题)
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u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29323 Accepted Submission(s): 13047
题目链接
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
要求用给出的公式输出当n=0到9的e的大小。
此题在于输出格式:╮(╯▽╰)╭。。。。。cout默认6位精度,而题目要求输出小数点后九位,所以要用到格式控制。。。。。。
╮(╯▽╰)╭。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。╮(╯▽╰)╭。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
代码:
#include <iostream>#include<stdio.h>#include<iomanip>using namespace std;long double f[10];void F(){ int i,j; int c=1; f[0]=1.0; for(i=1;i<=9;i++) { c=1; for(j=1;j<=i;j++) { c*=j; } f[i]=(long double)1.0/c; }}int main(){ int i,j; F(); long double s=0; cout<<"n e"<<endl; cout<<"- -----------"<<endl; for(i=0;i<=9;i++) { s+=f[i]; if(i>=3) cout<<setiosflags(ios::fixed)<<setprecision(9); cout<<i<<" "<<s<<endl; } return 0;}
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