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:
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
我的代码
#include <iostream>
#include <vector>
#include <iomanip>
#include <string>
#include <exception>
using namespace std;
bool preCheck(string a)//预处理函数
{int count=0;int c=0;for(int i=0;i<a.size();++i){if(a.at(i)=='.')//检验2.3.4这样的输入++count;if(count>=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<string> out;for(int i=0;i<count;i++){cin>>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<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;elsecout<<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