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
1 #include <stdio.h> 2 #include <algorithm> 3 #include <set> 4 #include <string.h> 5 #include <vector> 6 #include <math.h> 7 #include <queue> 8 #include <iostream> 9 #include <string> 10 using namespace std; 11 const int maxn = 1011; 12 int n; 13 double res=0; 14 bool isint(char c){ 15 if(c>='0' && c<='9')return true; 16 else return false; 17 } 18 bool isnum(string s){ 19 res=0; 20 if(s[0]!='-' && s[0]!='+' && (s[0]<'0' || s[0]>'9')) return false; 21 int flag=1; 22 if(s[0]=='-'){ 23 flag=-1; 24 s.erase(0,1); 25 } 26 else if(s[0]=='+'){ 27 s.erase(0,1); 28 } 29 int fp=0,np=0; 30 for(int i=0;i<s.length();i++){ 31 if(isint(s[i])){ 32 res=res*10+s[i]-'0'; 33 } 34 else if(s[i]=='.' && np==0){ 35 np++; 36 fp=i; 37 } 38 else{ 39 flag=0; 40 break; 41 } 42 } 43 if(flag==0)return false; 44 if(np==1){ 45 int xiao = s.length()-fp-1; 46 if(xiao>2) return false; 47 res = res / pow(10,xiao); 48 } 49 if(res>1000) return false; 50 //printf("%f %d ",res,flag); 51 res = res*flag; 52 //printf("%f %d ",res,flag); 53 return true; 54 } 55 56 int main(){ 57 scanf("%d",&n); 58 int cnt=0; 59 double total=0; 60 for(int i=1;i<=n;i++){ 61 string s; 62 cin>>s; 63 //res=0; 64 if(isnum(s)){ 65 cnt++; 66 total = total + res; 67 //printf("cnt: %d total: %f ",cnt,total); 68 } 69 else{ 70 printf("ERROR: %s is not a legal number ",s.c_str()); 71 } 72 } 73 if(cnt==0)printf("The average of 0 numbers is Undefined "); 74 else if(cnt!=1)printf("The average of %d numbers is %.2f ",cnt,total/cnt); 75 else printf("The average of %d number is %.2f ",cnt,total/cnt); 76 }
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