A1108. Finding Average
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 [-1000, 1000] 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 (<=100). 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<cstdio>
#include<iostream>
#include<string.h>
using namespace std;
char num[];
const double INF = 100000000.0;
double prase_(char num[]){
int len = strlen(num);
int pos = -, cnt = ;
int tag = ;
for(int i = ; num[i] != '\0'; i++){
if(!(num[i] >= '' && num[i] <= '' || num[i] == '.' || num[i] == '-')){
return INF;
}
if(num[i] == '.')
cnt++;
if(cnt > )
return INF;
}
if(num[] == '-')
tag = ;
for(pos = ; num[pos] != '.' && num[pos] != '\0'; pos++);
int P = , ansL = , ansR = ;
if(pos == '\0'){
for(int i = len - ; i >= tag; i--){
ansL += (num[i] - '') * P;
P *= ;
}
if(tag == )
return ansL * -1.0;
else return ansL * 1.0;
}else{
if(len - pos - > )
return INF;
P = ; ansL = ;
for(int i = pos - ; i >= tag; i--){
ansL += (num[i] - '') * P;
P *= ;
}
P = ; ansR = ;
for(int i = len - ; i > pos; i--){
ansR += (num[i] - '') * P;
P *= ;
}
double temp = ansR * 1.0;
for(int i = ; i < len - pos - ; i++){
temp *= 0.1;
}
if(tag == )
return ((double)ansL + temp) * -1.0;
else return (double)ansL + temp;
}
}
int main(){
int N, cnt = ;
double sum = , temp = ;
scanf("%d", &N);
for(int i = ; i < N; i++){
scanf("%s", num);
temp = prase_(num);
if(temp > 1000.0 || temp < -1000.0){
printf("ERROR: %s is not a legal number\n", num);
}else{
sum += temp;
cnt++;
}
}
if(cnt > ){
double prt = sum / (double)cnt;
printf("The average of %d numbers is %.2f\n", cnt, prt);
}else if (cnt == ){
printf("The average of 0 numbers is Undefined\n");
}else if(cnt == ){
double prt = sum / (double)cnt;
printf("The average of %d number is %.2f\n", cnt, prt);
}
cin >> N;
return ;
}
A1108. Finding Average的更多相关文章
- 【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 ...
- 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 ...
- PAT甲级——A1108 Finding Average【20】
The basic task is simple: given N real numbers, you are supposed to calculate their average. But wha ...
- 1108 Finding Average (20 分)
1108 Finding Average (20 分) The basic task is simple: given N real numbers, you are supposed to calc ...
- Pat1108: Finding Average
1108. Finding Average (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The b ...
- PAT 1108 Finding Average [难]
1108 Finding Average (20 分) The basic task is simple: given N real numbers, you are supposed to calc ...
- PAT_A1108#Finding Average
Source: PAT A 1108 Finding Average (20 分) Description: The basic task is simple: given N real number ...
- pat 1108 Finding Average(20 分)
1108 Finding Average(20 分) The basic task is simple: given N real numbers, you are supposed to calcu ...
- PAT (Advanced Level) 1108. Finding Average (20)
简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...
随机推荐
- vue2.0生命周期
https://www.cnblogs.com/goloving/p/8616989.html(copy )
- Python 基础之----网络编程
阅读目录 一 客户端/服务端架构 二 osi七层 三 socket层 四 socket是什么 五 套接字发展史及分类 六 套接字工作流程 七 基于TCP的套接字 八 基于UDP的套接字 九 粘包现象 ...
- python之路--基础数据类型的补充与深浅copy
一 . join的用法 lst =['吴彦祖','谢霆锋','刘德华'] s = '_'.join(lst) print(s) # 吴彦祖_谢霆锋_刘德华 # join() "*" ...
- Django restframework之Token验证的缺陷及jwt的简单使用
一.主要缺陷: 1.Token验证是放在一张表中,即authtoken_token中,key没有失效时间,永久有效,一旦泄露,后果不可想象,安全性极差. 2.不利于分布式部署或多个系统使用一套验证,a ...
- Java语言支持的3种变量类型
类变量(静态变量):独立于方法之外的变量,用 static 修饰. 实例变量(全局变量):独立于方法之外的变量,不过没有 static 修饰. 局部变量:类的方法中的变量. 例子如下: public ...
- Delphi调用MSSQL存储过程返回的多个数据集的方法
varaintf:_Recordset;RecordsAffected:OleVariant; begin ADOStoredProc1.Close;ADOStoredProc1.Open;aintf ...
- delphi 子窗体只能最小化不能关闭的解决方案
cnpack下载地址:http://www.cnpack.org/showdetail.php?id=726&lang=zh-cn 时候创建的子窗体不能关闭,点关闭按钮时子窗体最小化了. 出现 ...
- JAVA 变量 数据类型 运算符 知识小结
---------------------------------------------------> JAVA 变量 数据类型 运算符 知识小结 <------------------ ...
- Git要点
前面的话 本文将总结Git要点 版本管理工具 [作用] 1.备份文件 2.记录历史 3.回到过去 4.对比差异 [分类] 1.手动版本控制(又叫人肉VCS) 2.LVCS 本地 3.CVCS 集中式( ...
- ATM实验感受
public class Account { private String accountID; private String accountname; private String operated ...