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:

#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:

#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;
}