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
- Unit 1.前端基础之html
一.什么是html 定义:全称是超文本标记语言(HyperText Markup Language),它是一种用于创建网页的标记语言.标记语言是一种将文本(Text)以及文本相关的其他信息结合起来,展 ...
- 图解Python的直接赋值与浅拷贝和深度拷贝三者区别
直接赋值:其实就是对象的引用(别名). 浅拷贝(copy):拷贝父对象,不会拷贝对象的内部的子对象. 深拷贝(deepcopy): copy 模块的 deepcopy 方法,完全拷贝了父对象及其子对象 ...
- 4面向对象(OOP)
学习线路 初学: 熟悉语法 进阶: 1.23种设计模式 2.6中开发原则 高级: 1.优化 2.架构 3.安全 概念 类:一类具有相同特性的事物的抽象描述,用一个java类表示. 成员变量:抽取的属性 ...
- maven 中的pom中的 dependencyManagement 和 dependencies
参考:maven pom.xml 中 dependencyManagement和dependencies详解 现在的项目基本上都是使用多module来管理的,这就涉及到一个问题,多module之间如何 ...
- jdbc一点小笔记
JDBC的常用接口的步骤, 1使用Driver或者Class.forName()进行注册驱动: 2使用DriverManager进行获取数据库的链接.使用Connection获取语句对象.使用语句对象 ...
- elasticsearch索引合并
参考地址:http://cwiki.apachecn.org/display/Elasticsearch/Reindex+API 1.首先插入准备数据,创建两个索引. (1).PUT http:// ...
- JQuery跳出each循环的方法(包含数组遍历)
0. 前言 也许我们通过 jquery 的循环方法进行数组遍历,但是当不符合条件时,怎么跳出当前循环?(即用each方法内,当不满足条件时想break跳出循环体,想continue继续执行下一个循环遍 ...
- 1.rabbitmq高可用方案
采用标准集群模式 HAPROXY + rabbitmq 2个 ram 和 一个 disk 节点 主机规划: 192.168.157.128 haproxy keepalive 主 ram节点 1 ...
- 前端动态属性页面的 要用id做name 因为这样方便在提交表单时候取到值
前端动态属性页面的 要用id做name 因为这样方便在提交表单时候取到值
- Nginx 缓存针对打开的文件句柄与原文件信息
L:108 open_file_cache syntax: open_file_cache off; open_file_cache max=N[inactive=time](inactive表示 ...