題目信息 1108. Finding Average (20)

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 1:75 -3.2 aaa 9999 2.3.4 7.123 2.35Sample Output 1:ERROR: aaa is not a legal numberERROR: 9999 is not a legal numberERROR: 2.3.4 is not a legal numberERROR: 7.123 is not a legal numberThe average of 3 numbers is 1.38Sample Input 2:2aaa -9999Sample Output 2:ERROR: aaa is not a legal numberERROR: -9999 is not a legal numberThe average of 0 numbers is Undefined

我的代碼

#include <iostream>
#include <vector>
#include <iomanip>
#include <string>
#include <exception>
using namespace std;
bool preCheck(string a)//預處理函數(shù)
{ 
   int count=0;
   int c=0; 
   for(int i=0;i<a.size();++i)
   {
     if(a.at(i)=='.')//檢驗2.3.4這樣的輸入 
     ++count; 
     if(count>=1)//檢測decimal places>2的情況 
        ++c; } 
    if(count>1) 
      return false; if(c>3) return false; 
    return true;
}
int main()
{
   int count; 
   int size=0; 
   double sum=0; 
   double itmp=0; 
   cin>>count; 
   string *tmp=new string[count]; 
   vector<string> out; 
   for(int i=0;i<count;i++)
   { 
      cin>>tmp[i]; 
      try { 
              if(!preCheck(tmp[i]))//由于stod()能轉(zhuǎn)化2.3.4這樣的數(shù)，所以進行了預處理 
              throw exception(); 
              itmp=stod(tmp[i]); 
              if(itmp>1000||itmp<-1000)//題目要求 
                  throw exception(); 
              ++size; 
              sum+=itmp;   
       }catch(exception &e)//由于stod轉(zhuǎn)化失敗后會發(fā)出異常，所以統(tǒng)一將不正確的輸入進行異常處理 
       { 
              out.push_back(tmp[i]); 
       } 
 } 
 for(int i=0;i<count-size;i++) 
 { 
      cout<<"ERROR: "<<out.at(i)<<" is not a legal number"<<endl; 
 }
 if(size!=0) { 
      sum=sum/size; 
      if(size==1) 
        cout<<fixed<<setprecision(2)<<"The average of "<<size<<" number is "<<sum<<endl; 
      else 
        cout<<fixed<<setprecision(2)<<"The average of "<<size<<" numbers is "<<sum<<endl; }
  else { cout<<"The average of 0 numbers is Undefined"<<endl;  } 
 return 0;
}

遇到的問題

• 如何將string轉(zhuǎn)化為double
• 3.這樣的輸入是正確的
• 只有一個正常輸入時，此時的輸出是number