1054. 求平均值 (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
CHEN, Yue
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1:7 5 -3.2 aaa 9999 2.3.4 7.123 2.35输出样例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输入样例2:
2 aaa -9999输出样例2:
ERROR: aaa is not a legal number ERROR: -9999 is not a legal number The average of 0 numbers is Undefined
01:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 | #include <cstring> #include <iostream> #include <cstdio> using namespace std; int main(){ int n; string s; double sum = 0; int count = 0; cin >> n; for ( int k = 0; k < n; k++){ cin >> s; double b = 0, c = 0; //b记录整数部分,c记录小数部分 if (s[0] == '-' ){ //为负数的情况 int i = 1; while (s[i] >= '0' && s[i] <= '9' ){ b = b * 10 + s[i] - '0' ; i++; } if (s[i] == '.' ){ int j = i + 1; double l = 0.1; while (s[j] >= '0' && s[j] <= '9' ){ c += l * (s[j] - '0' ); j++; l *= 0.1; } if (j - i - 1 > 2 || j + 1 < s.length()){ cout << "ERROR: " << s << " is not a legal number" << endl; continue ; } } if (b + c < -1000 || b + c> 1000){ cout << "ERROR: " << s << " is not a legal number" << endl; continue ; } count++; sum += -b - c; } else if (s[0] >= '0' && s[0] <= '9' ){ //非负数 int i = 0; while (s[i] >= '0' && s[i] <= '9' ){ b = b * 10 + s[i] - '0' ; i++; } if (s[i] == '.' ){ int j = i + 1; double l = 0.1; while (s[j] >= '0' && s[j] <= '9' ){ c += l * (s[j] - '0' ); j++; l *= 0.1; } if (j - i - 1 > 2 || j < s.length()){ cout << "ERROR: " << s << " is not a legal number" << endl; continue ; } } if (b + c < -1000 || b + c> 1000){ cout << "ERROR: " << s << " is not a legal number" << endl; continue ; } count++; sum += b + c; } else { cout << "ERROR: " << s << " is not a legal number" << endl; } } if (count == 0){ printf ( "The average of 0 numbers is Undefined" ); } else if (count == 1) printf ( "The average of 1 number is %.2lf" , sum); else printf ( "The average of %d numbers is %.2lf" , count, sum / count); return 0; } |
02:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 | #include <iostream> #include <cstring> #include <cstdio> using namespace std; int main(){ char a[50], b[50]; //尽量大点,不能提前预知非法数据的长短 int n, count = 0, flag = 0; double sum, temp; cin >> n; for ( int i = 0; i < n; i++){ flag = 0; cin >> a; sscanf (a, "%lf" , &temp); //a必须是char类型,不能是string sprintf (b, "%.2lf" , temp); for ( int j = 0; j < strlen (a); j++){ if (a[j] != b[j]) flag = 1; } if (flag == 1 || temp > 1000 || temp < -1000){ cout << "ERROR: " << a << " is not a legal number" << endl; continue ; } sum += temp; count++; } if (count == 0){ printf ( "The average of 0 numbers is Undefined" ); } else if (count == 1) printf ( "The average of 1 number is %.2lf" , sum); else printf ( "The average of %d numbers is %.2lf" , count, sum / count); return 0; } |
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