PAT 乙级 (Basic Level) Practice (中文)1054
2019-03-01 17:05
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1054 求平均值 (20 分)
本题的基本要求非常简单:给定 NNN 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 NNN(≤100)。随后一行给出 NNN 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
代码
//参考自https://www.liuchuo.net/archives/617 #include <cstdio> #include <cstring> #include <iostream> using namespace std; int main () { int N, count = 0; double temp, sum = 0.0; char st[10], e[10]; //原始字符串,转换后的字符串 scanf ("%d", &N); for (int i = 0; i < N; i++) { scanf ("%s", st); sscanf (st, "%lf", &temp); sprintf (e, "%.2f", temp); int flag = 0; //正常实数 for (int j = 0; j < strlen(st); j++) { if (st[j] != e[j]) { //大为惊叹,竟然可以这样判断 flag = 1; break; } } if (flag || temp > 1000 || temp < -1000) { //非法情况 printf ("ERROR: %s is not a legal number\n", st); continue; } else { sum += temp; count++; } } if (count == 0) { printf ("The average of 0 numbers is Undefined\n"); } else if (count == 1) { printf ("The average of 1 number is %.2f\n", sum); } else { printf ("The average of %d numbers is %.2f\n", count, sum / count); } return 0; }
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