本题的基本要求非常简单:给定 N 个实数,计算它们的平均值。但复杂的是有些输入数据可能是非法的。一个“合法”的输入是 [−1000,1000] 区间内的实数,并且最多精确到小数点后 2 位。当你计算平均值的时候,不能把那些非法的数据算在内。
输入格式:
输入第一行给出正整数 N(≤100)。随后一行给出 N 个实数,数字间以一个空格分隔。
#### 输出格式:
对每个非法输入,在一行中输出 ERROR: X is not a legal number,其中 X 是输入。最后在一行中输出结果:The average of K numbers is Y,其中 K 是合法输入的个数,Y 是它们的平均值,精确到小数点后 2 位。如果平均值无法计算,则用 Undefined 替换 Y。如果 K 为 1,则输出 The average of 1 number is Y。
#### 输入样例 1:
1 2
7 5 -3.2 aaa 9999 2.3.4 7.123 2.35
#### 输出样例 1:
1 2 3 4 5
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
#### 输入样例 2:
1 2
2 aaa -9999
#### 输出样例 2:
1 2 3
ERROR: aaa is not a legal number ERROR: -9999 is not a legal number The average of 0 numbers is Undefined
boolcheck(string t, double* x){ int dot = -1; int dotNum = 0; bool ok = true; int i = 0; if (t[i] == '-')i++; for ( ; i < t.length(); i++) { if (!(t[i] >= '0' && t[i] <= '9')) { if (t[i] == '.') { dotNum++; if (dotNum == 1) dot = i; } else { ok = false; break; } } } if (dotNum > 1) ok = false; if (dot != -1 && t.length() - dot - 1 > 2)ok = false; if (ok) { *x = atof(t.data()); if (fabs(*x) > 1000) ok = false; } return ok; }
intmain(){ int n; cin >> n; int count = 0; double sum = 0; string str; for (int j = 0; j < n; j++) { double StrToNum = 0; cin >> str; if (check(str, &StrToNum)) { sum += StrToNum; count++; } else { cout << "ERROR: " << str << " is not a legal number" << endl; } }
if (count == 1) { cout << "The average of " << count << " number is " << fixed << setprecision(2) << sum / count << endl; } elseif (count == 0) { cout << "The average of 0 numbers is Undefined" << endl; } else { cout << "The average of " << count << " numbers is " << fixed << setprecision(2) << sum / count << endl; } }