The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−] and is accurate up to no more than 2 decimal places. When you calculate the average, those illegal numbers must not be counted in.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤). Then N numbers are given in the next line, separated by one space.

Output Specification:

For each illegal input number, print in a line ERROR: X is not a legal number where X is the input. Then finally print in a line the result: The average of K numbers is Y where K is the number of legal inputs and Y is their average, accurate to 2 decimal places. In case the average cannot be calculated, output Undefined instead of Y. In case K is only 1, output The average of 1 number is Y instead.

Sample Input 1:

7
5 -3.2 aaa 9999 2.3.4 7.123 2.35

Sample Output 1:

ERROR: aaa is not a legal number
ERROR: 9999 is not a legal number
ERROR: 2.3.4 is not a legal number
ERROR: 7.123 is not a legal number
The average of 3 numbers is 1.38

Sample Input 2:

2
aaa -9999

Sample Output 2:

ERROR: aaa is not a legal number
ERROR: -9999 is not a legal number
The average of 0 numbers is Undefined
 #include <iostream>
#include <vector>
using namespace std;
#define inf 1000
double sum = 0.0, number;
int n, nums = ;
int main()
{
cin >> n;
char str1[], str2[];
while (n--)
{
cin >> str1;
bool flag = true;
sscanf(str1, "%lf", &number);//将字符转换为浮点数值
sprintf(str2, "%0.2f", number);//按要求输出
for (int i = ; str1[i] != '\0'&& flag; ++i)
if (str1[i] != str2[i])
flag = false;
if (!flag || number<-inf || number>inf)
cout << "ERROR: " << str1 << " is not a legal number" << endl;
else
{
nums++;
sum += number;
}
}
cout << "The average of " << nums << (nums == ? " number is " : " numbers is ");
if (nums == )
cout << "Undefined" << endl;
else
printf("%0.2f\n", sum / nums);
return ;
}

PAT甲级——A1108 Finding Average【20】的更多相关文章

  1. PAT甲级——1108.Finding Average (20分)

    The basic task is simple: given N real numbers, you are supposed to calculate their average. But wha ...

  2. 【刷题-PAT】A1108 Finding Average (20 分)

    1108 Finding Average (20 分) The basic task is simple: given N real numbers, you are supposed to calc ...

  3. PAT Advanced 1108 Finding Average (20 分)

    The basic task is simple: given N real numbers, you are supposed to calculate their average. But wha ...

  4. Day 007:PAT训练--1108 Finding Average (20 分)

    话不多说: 该题要求将给定的所有数分为两类,其中这两类的个数差距最小,且这两类分别的和差距最大. 可以发现,针对第一个要求,个数差距最小,当给定个数为偶数时,二分即差距为0,最小:若给定个数为奇数时, ...

  5. PAT A1108 Finding Average (20 分)——字符串,字符串转数字

    The basic task is simple: given N real numbers, you are supposed to calculate their average. But wha ...

  6. 【PAT甲级】1108 Finding Average (20分)

    题意: 输入一个正整数N(<=100),接着输入一行N组字符串,表示一个数字,如果这个数字大于1000或者小于1000或者小数点后超过两位或者压根不是数字均为非法,计算合法数字的平均数. tri ...

  7. PAT (Advanced Level) 1108. Finding Average (20)

    简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...

  8. PAT甲题题解-1108. Finding Average (20)-字符串处理

    求给出数的平均数,当然有些是不符合格式的,要输出该数不是合法的. 这里我写了函数来判断是否符合题目要求的数字,有点麻烦. #include <iostream> #include < ...

  9. PAT 甲级 1035 Password (20 分)

    1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for th ...

随机推荐

  1. Java中获取前一天和后一天时间

    今天在开发项目的时候遇到一个问题就是怎么获取当前时间的前一天和后一天,这个实现的逻辑并不复杂,自己要写的话的也不是难事,但是貌似感觉没必要自己写这样的方法,想想Java中的Calendar类应该有这样 ...

  2. 模拟+算贡献——cf1195D

    比赛的时候没看到模数,用java大数在写,最后看到的时候已经慌了.. 把贡献算清楚就可以 下面是贡献的推导 有五位数 abcde * 10个 有两位数 fg * 3 个 那么这两种数组成的情况就是 a ...

  3. NX二次开发-UFUN设置工程图PNG图片高度UF_DRF_set_image_height

    #include <uf.h> #include <uf_drf.h> UF_initialize(); //插入PNG char* file_name = "D:\ ...

  4. linux基本命令vim

    拷贝当前行 yy,拷贝当前行向下的5行  5yy, 并粘贴(p). 删除当前航  dd,删除当前行向下的5行 5dd. 在文件中查找某个单词[命令行下/关键字,回车查找, 输入n 就是查找下一个] 查 ...

  5. 在linux中的rpm -ivh 是干什么的呢?

    在linux中的rpm -ivh 是干什么的呢?   RMP 是 LINUX 下的一种软件的可执行程序,你只要安装它就可以了.这种软件安装包通常是一个RPM包(Redhat Linux Packet ...

  6. docker搭建jenkins

    这里是在linux环境下安装docker之后,在doucer内安装jenkins --------------------docker 安装 jenkins---------------------- ...

  7. [转]ThinkPHP分页实例

    很多人初学thinkphp时,不太熟悉thinkphp的分页使用方法,现在将自己整理的分页方法分享下,有需要的朋友可以看看.   控制器中的代码:   $db = M("cost" ...

  8. python之tkinter学习目录

    前言 下面的目录结构,采用的学习视频资料是网易云课堂中[莫凡]老师的,在目录的最下面的地方给出了对应的链接! 学习是逐渐积累起来的,代码也是!下面的每一篇中的对应代码,都秉承着这样的一个理念:代码是成 ...

  9. Nginx:413 Request Entity Too Large 的解决方法

    报错信息413 Request Entity Too Large 解决方法: (20M大小,自己调节,根据文件大小.)修改 php 的配置文件 /etc/php5/fpm/php.ini upload ...

  10. 实时查询系统架构:spark流式处理+HBase+solr/ES查询

    最近要做一个实时查询系统,初步协商后系统的框架 1.流式计算:数据都给spark 计算后放回HBase 2.查询:查询采用HBase+Solr/ES