1108. Finding Average (20)

题目信息 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

我的代码

#include 
#include 
#include 
#include 
#include 
using namespace std;
bool preCheck(string a)//预处理函数
{ 
   int count=0;
   int c=0; 
   for(int i=0;i=1)//检测decimal places>2的情况 
        ++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 out; 
   for(int i=0;i>tmp[i]; 
      try { 
              if(!preCheck(tmp[i]))//由于stod()能转化2.3.4这样的数，所以进行了预处理 
              throw exception(); 
              itmp=stod(tmp[i]); 
              if(itmp>1000||itmp<-1000)//题目要求 
                  throw exception(); 
              ++size; 
              sum+=itmp;   
       }catch(exception &e)//由于stod转化失败后会发出异常，所以统一将不正确的输入进行异常处理 
       { 
              out.push_back(tmp[i]); 
       } 
 } 
 for(int i=0;i

遇到的问题

• 如何将string转化为double
• 3.这样的输入是正确的
• 只有一个正常输入时，此时的输出是number