HDU 1012.u Calculate e【水】【8月16】

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;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

时间: 2024-11-05 02:38:20

HDU 1012.u Calculate e【水】【8月16】的相关文章

hdu 1012:u Calculate e(数学题,水题)

u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28686    Accepted Submission(s): 12762 Problem Description A simple mathematical formula for e iswhere n is allowed to go to infinit

poj 1517 &amp; hdu 1012 u Calculate e(简单阶乘)

POJ链接 :http://poj.org/problem?id=1517 HDU链接:http://acm.hdu.edu.cn/showproblem.php?pid=1012 Description A simple mathematical formula for e is e=Σ0<=i<=n1/i! where n is allowed to go to infinity. This can actually yield very accurate approximations o

水题/hdu 1012 u Calculate e

题意 求n=0~9时的sigma(1/n!) 分析 因为刚学c++ 所以对浮点操作还是很不熟练,正好来了这么一道题 Accepted Code 1 /* 2 PROBLEM:hdu 1012 3 AUTHER:Nicole Lam 4 MEMO:水题 5 */ 6 #include<iostream> 7 #include<iomanip> 8 using namespace std; 9 double a[10]; 10 int main() 11 { 12 cout<&l

HDU 1012 u Calculate e(简单阶乘计算)

传送门: 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): 52607    Accepted Submission(s): 24106 Problem Description A simple mathematical

HDU 1012 u Calculate e【暴力打表,水】

u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46844    Accepted Submission(s): 21489 Problem Description A simple mathematical formula for e is where n is allowed to go to infini

Hdu 1012 u Calculate e

u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46276    Accepted Submission(s): 21237 Problem Description A simple mathematical formula for e is where n is allowed to go to infini

hdu 1251 统计难题 (map水过)

# include <stdio.h> # include <algorithm> # include <string.h> # include <map> # include <iostream> using namespace std; int main() { char a; string x; map<string,int>q; while(true) { scanf("%c",&a); if(a=

HDU 4893 Wow! Such Sequence! 水线段树

思路: 线段树走起.. 写完这题就退役T^T 单点更新的时候直接找到这个点的最近fib,然后维护当前和 和 fib的和 #include<stdio.h> #include<string.h> #include<iostream> #include<math.h> #include<algorithm> #include<queue> #include<map> #include<set> #include&l

hdu 1999 不可摸数 水题。

不可摸数 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7966    Accepted Submission(s): 2024 Problem Description s(n)是正整数n的真因子之和,即小于n且整除n的因子和.例如s(12)=1+2+3+4+6=16.如果任何数m,s(m)都不等于n,则称n为不可摸数. Input 包