PAT日志 1108

顽强的小白

1108 Finding Average （20 分）

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

题目解析

处理字符串的问题，只要合法的数字。

思路：

• 首先判断字符串里面是不是只有数字、负号和小数点
• 再来判断是不是有小数点，是不是只有一个
• 如果有小数点，判断小数点后是不是不超过两位，同时要把字符串转换成数字
• 最后判断数字是不是在给定范围内

注意点：

• 小数点可以在最后，例如 89.
• 没有 ‘+’号的例子
• 注意只有一个合法数字的输出是单数的 number 但是仍然要用补齐小数点后两位的方法输出
• 注意没有合法数字的输出

代码实现

#include <cstdio>
#include <iostream>   //它和下面这个一般是配套使用
#include <string>   //为了用 string 的 find 函数 没有用字符串  
#include <algorithm>
using namespace std;
const int maxn=105;
double sum=0;
int cnt=0;
bool judge(string now){
 for(int i=0;i<now.size();++i){ //先判断是不是只有合法的元素，过滤掉字母
  if((now[i]<='9'&&now[i]>='0')||now[i]=='-'||now[i]=='.'){
   continue; 
  }else return false;
  
 }
 int i=0;
 if(now[0]=='-') i++;
 int front=0,rear=0,pos;
 double num=0;
 if(now.find('.')!=string::npos ){ //找到一个小数点 
  pos=now.find('.',i);
  if(now.find('.',pos+1)!=string::npos){ //不止包含一个小数点 
   return false;
  }
     if(now.size()-1-pos>2) return false; //判断小数点后的位数 
     for(int j=i;j<pos;++j){
      front*=10; 
   front+=now[j]-'0';
  }
  for(int j=pos+1;j<now.size();++j){
   rear*=10;
   rear+=now[j]-'0';
  }
  if(now.size()-1-pos==1){
   num=front+rear/10.0;
  }else if(now.size()-1-pos==2){
   num=front+rear/100.0;
  }else if(now.size()-1-pos==0){
   num=front+rear;
  }
  
 }else{ //如果没有小数点，整数 
  for(int j=i;j<now.size();++j){
   num*=10;
   num+=now[j]-'0'; 
  }
 } 
 if(now[0]=='-') num=-num;
 if(num<=1000&&num>=-1000){
  //printf("%.2lf\n",num);
  sum+=num;
  cnt++;
  return  true; 
 }
 else 
  return false;
}
int main(){
 int n;
 scanf("%d",&n);
 string now;
 for(int i=0;i<n;++i){
  cin>>now;
  if(judge(now)==false){
   cout<<"ERROR: "<<now<<" is not a legal number"<<endl;
  }
 }
 if(cnt>1)
  printf("The average of %d numbers is %.2lf\n",cnt,sum/cnt);
 else if(cnt==1){
  printf("The average of 1 number is %.2lf\n",sum);
 }else
  printf("The average of 0 numbers is Undefined\n");
 return 0;
}