1054 求平均值

这道题很好，可以作为一个简便方法，即判断基本数据类型的某数是否合法，值得回顾。

#include<iostream>
#include<cstring>
using namespace std;

char str1[5000],str2[5000];
int main() {
    int n,cnt = 0;
    double a,sum = 0;
    cin>>n;
    while(n--) {
        scanf("%s",str1);
        sscanf(str1,"%lf",&a);
        sprintf(str2,"%.2f",a);
        bool flag = true;
        for(int i = 0; i < strlen(str1); ++i) {
            //假设str1为2.333,str2为2.33,那么 str1与str2两值不相等
            //假设str1为2.3,  str2为2.30,那么 str1与str2两值相等 
            if(str1[i] != str2[i]) flag = false;
        }
        if(flag == false || a < -1000 || a > 1000)
            printf("ERROR: %s is not a legal number\n",str1);
        else {
            sum += a;
            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);
    return 0;
}