PAT A1108 Finding Average （20 分）——字符串，字符串转数字

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

 #include <stdio.h>
 #include <algorithm>
 #include <set>
 #include <string.h>
 #include <vector>
 #include <math.h>
 #include <queue>
 #include <iostream>
 #include <string>
 using namespace std;
 const int maxn = ;
 int n;
 double res=;
 bool isint(char c){
     if(c>='' && c<='')return true;
     else return false;
 }
 bool isnum(string s){
     res=;
     if(s[]!='-' && s[]!='+' && (s[]<'' || s[]>'')) return false;
     int flag=;
     if(s[]=='-'){
         flag=-;
         s.erase(,);
     }
     else if(s[]=='+'){
         s.erase(,);
     }
     int fp=,np=;
     for(int i=;i<s.length();i++){
         if(isint(s[i])){
             res=res*+s[i]-'';
         }
         else if(s[i]=='.' && np==){
             np++;
             fp=i;
         }
         else{
             flag=;
             break;
         }
     }
     if(flag==)return false;
     if(np==){
         int xiao = s.length()-fp-;
         if(xiao>) return false;
         res = res / pow(,xiao);
     }
     if(res>) return false;
     //printf("%f %d\n",res,flag);
     res = res*flag;
     //printf("%f %d\n",res,flag);
     return true;
 }

 int main(){
     scanf("%d",&n);
     int cnt=;
     double total=;
     for(int i=;i<=n;i++){
         string s;
         cin>>s;
         //res=0;
         if(isnum(s)){
             cnt++;
             total = total + res;
             //printf("cnt: %d total: %f\n",cnt,total);
         }
         else{
             printf("ERROR: %s is not a legal number\n",s.c_str());
         }
     }
     if(cnt==)printf("The average of 0 numbers is Undefined\n");
     else if(cnt!=)printf("The average of %d numbers is %.2f\n",cnt,total/cnt);
     else printf("The average of %d number is %.2f\n",cnt,total/cnt);
 }

注意点：字符串转数字的判断，按条件一个个判断就好了，注意输入输出时的double要写对。测试点3是只有一个数的情况，1个数时number没有s，没注意所以错了