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HDOJ题目地址：传送门

u Calculate e

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.

n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<memory.h>
using namespace std;
int jiecheng(int m){
    if(m<=1)
        return 1;
    else
        return m*jiecheng(m-1);
}
int main(){
    printf("n e\n- -----------\n");
    for(int i=0;i<=9;i++){
        double result=0;
        for(int j=0;j<=i;j++){
            result+=(1.0/jiecheng(j));
        }
        if(i==0||i==1)
            printf("%d %.0lf\n",i,result);
        else if(i==2)
            printf("%d %.1lf\n",i,result);
        else
            printf("%d %.9lf\n",i,result);
    }
}

ACM--公式--HDOJ 1012--u Calculate e--水

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原文地址：http://blog.csdn.net/qq_26891045/article/details/51881838