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 [−] 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 (≤). 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
乙级真题
#include <iostream> #include <cstring> using namespace std; int main() { int N;cin>>N; double avg=0; int val,coun=0; double a; while(N--){ bool right=true; char tmp[50],tmp2[50]; scanf("%s",tmp); sscanf(tmp,"%lf",&a); sprintf(tmp2,"%.2f",a); for(int i=0;i<strlen(tmp);i++) if(tmp[i]!=tmp2[i]) right=false; if(right&&a<=1000&&a>=-1000) { avg+=a; coun++; } else printf("ERROR: %s is not a legal number ",tmp); } if(coun!=0) { if(coun>1){ avg/=coun; printf("The average of %d numbers is %.2f",coun,avg); }else printf("The average of 1 number is %.2f",avg); }else printf("The average of 0 numbers is Undefined"); system("pause"); return 0; }