1108 Finding Average（20 分）（C++）

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 Undefinedinstead 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

解题思路：

atof：将数字字符串转为浮点数，但是atof是针对char*类型。string转为char*，使用s.c_str()

stof：将string类型转为浮点数float（本题可以直接用这个）

stod: 将string转为double类型

#include <iostream>
#include <string>
using namespace std;
int main(){
	int n, cnt = 0;
	double sum ,temp;
	string s;
	scanf("%d", &n);
	for(int i = 0; i < n; ++ i){
		bool flag = true, tempflag = true;
		cin >> s;
		for(int j = 0; j < s.length(); ++ j){
			if(j == 0 && (s[j] == '-' || s[j] == '+'))
				continue;
			if(s[j] == '.' && tempflag){
				tempflag = false;
				if(s.length() - j > 3)
					flag = false;
			}
			else if(s[j] > '9' || s[j] < '0')
				flag = false;
		}
		if(flag){
			temp = stod(s);
			if(temp > 1000 || temp < -1000)
				flag = false;
		}
		if(!flag)
			printf("ERROR: %s is not a legal number\n", s.c_str());
		else{
			sum += temp;
			++ cnt;
		}
	}
	if(cnt == 0)
        printf("The average of 0 numbers is Undefined");
  else if(cnt == 1)
        printf("The average of 1 number is %.2f",sum);
  else
        printf("The average of %d numbers is %.2f",cnt,sum/cnt);
}

方法二：手动实现数字字符串转浮点数

#include<iostream>
#include<cstdio>
#include<string>
#include<cctype>
#include<cmath> 
using namespace std;
int main(){
    int n,cnt=0;
    double sum=0.0;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        string s;
        cin>>s;
        bool flag=0,flag2=0;
        int u=0;
        double temp=0.0;
        for(int j=0;j<s.length();j++){
        	if(j==0&&s[0]=='-'){
        		flag2=1;
        		continue;
			}
            if(!isdigit(s[j])&&(s[j]!='.'||u!=0)){
                flag=1;
                break;
            }
            if(s[j]=='.')
                u=j;
            else if(isdigit(s[j])){
                if(u==0)
                    temp=temp*10+(s[j]-'0');
                else{
                    if(j-u>2){
                        flag=1;
                        break;
                    }
                    else
                        temp=temp+(s[j]-'0')*pow(10,u-j);
                }
            }
        }
        if(temp>1000)
        	flag=1;
        if(flag==1)
            cout<<"ERROR: "<<s<<" is not a legal number"<<endl; 
        else {
        	if(flag2==0){
        		sum+=temp;
        		cnt++;
			}
            else {
            	sum-=temp;
                cnt++;
            }
        }
    }
    if(cnt==1)
       printf("The average of %d number is ",cnt);
    else
       printf("The average of %d numbers is ",cnt);
    if(cnt==0)
        printf("Undefined");
    else
        printf("%.2f\n",sum/cnt);
}