The basic requirement of this question is very simple: Given N real numbers, calculate their average value. But complex is that some input data may be illegal. A "legitimate" input is a real number in the interval [1000,1000] and is at most accurate to two decimal places. When you calculate the average, you can't count the illegal data.

Input format:

Input the first line to give a positive integer N (< 100). The next line gives N real numbers separated by a space.

Output format:

For each illegal input, ERROR: X is not a legal number, where X is the input. Finally, The average of K numbers is Y, where K is the number of legitimate inputs and Y is their average value, accurate to two decimal places. If the average cannot be calculated, replace Y with Undefined. If K is 1, The average of 1 number is Y.

Input Sample 1:

7
5 -3.2 aaa 9999 2.3.4 7.123 2.35

Output Sample 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

Input Sample 2:

2
aaa -9999

Output Sample 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<stdlib.h>

int main()
{
	int N = 0;
	scanf("%d",&N);
	char str[100] = {};	//Original string
	char dest[100] = {}; //Keep two decimal strings
	double num = 0;
	int flag = 0;		//Is the record illegal?
	double Y = 0;		
	int K = 0;		
	for(int i = 0 ; i < N ; i++)
	{
		flag = 0;
		scanf("%s",str);
		sscanf(str,"%lf",&num);
		sprintf(dest,"%.2f",num);
		for(int i = 0 ;'\0' != str[i];i++)
		{
			if(str[i] != dest[i])
			{
				flag = 1;
				break;
			}
		}
		if(1 == flag || num < -1000 || num > 1000)
			printf("ERROR: %s is not a legal number\n",str);
		else
		{
			Y += num;
			K++;
		}
	}
	if(0 == K)
		printf("The average of 0 numbers is Undefined\n");
	else if(1 == K)
		printf("The average of 1 number is %.2f\n",Y);
	else
		printf("The average of %d numbers is %.2lf\n",K,Y/K);
	return 0;
}