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

注意点

  • 字符串数组不能开太小，否则会造成 “运行时错误”，即数组越界。实测大小至少为50.

方法一

来自柳神，不容易想到。运用 sscanf 和 sprintf 后可以方便判断是否为合法数字。利用 sscanf(a, “%lf”, &temp) 将原字符串输入为 double 型，再利用 sprintf(b, “%.2f”,temp) 将该 double 型数字输出为保留两位小数的字符串。之后判断保留两位小数后的字符串和原字符串是否相同，即可判断是否合法。

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int main() {
    int n, cnt = 0;
    char a[50], b[50];
    double temp, sum = 0.0;
    cin >> n;
    for(int i = 0; i < n; i++) {
        scanf("%s", a);
        sscanf(a, "%lf", &temp);
        sprintf(b, "%.2f",temp);
        int flag = 0;
        for(int j = 0; j < strlen(a); j++)
            if(a[j] != b[j]) flag = 1;
        if(flag || temp < -1000 || temp > 1000) {
            printf("ERROR: %s is not a legal number\n", a);
            continue;
        } else {
            sum += temp;
            cnt++;
        }
    }
    if(cnt == 1)
        printf("The average of 1 number is %.2f", sum);
    else if(cnt > 1)
        printf("The average of %d numbers is %.2f", cnt, sum / cnt);
    else
        printf("The average of 0 numbers is Undefined");
    return 0;
}

方法二

我的方法比较笨，但最直接，思路简单，就是直接遍历原字符串的每个字符，判断是否合法。

#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

int main() {
    int n, cnt = 0;
    scanf("%d", &n);
    char str[50];
    double total = 0;
    for (int i = 0; i < n; i++) {
        scanf("%s", str);
        int len = strlen(str), index = -1;
        bool flag = true;
        for (int i = 0; i < len; i++) {
            if (str[i] == '.') {
                if (i < len - 3 || index != -1) {
                    flag = false;
                    break;
                }
                index = i;
            } else if ((str[i] < '0' || str[i] > '9') && !(i == 0 && str[i] == '-')) {
                flag = false;
            }
        }
        if (!flag) {
            printf("ERROR: %s is not a legal number\n", str);
        } else {
            double num = stod(str);
            if (num < -1000 || num > 1000) {
                printf("ERROR: %s is not a legal number\n", str);
            } else {
                cnt++;
                total += num;
            }
        }
    }
    if (cnt == 0) {
        printf("The average of 0 numbers is Undefined");
    } else if (cnt == 1) {
        printf("The average of 1 number is %.2f", total / cnt);
    } else {
        printf("The average of %d numbers is %.2f", cnt, total / cnt);
    }
    return 0;
}