PTA 1108 Finding Average (20分)

1108 Finding Average (20分)

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

 

题目大意：

以字符串的形式输入n个数字串，统计出其中合法的数字串的平均值。

合法的数字串满足的条件

1. 不包含除了数字和"-"以及"."以外的字符
2. 小数点的个数不能大于1个，并且小数点可以出现在数字串末尾，但不能出现在数字串首
3. "-"只能出现在字符串首位
4. 对于浮点数，它的小数位不能超过2位
5. 数字的大小在[-1000,1000]之内

注意事项：

1.

若最终求出的合法数字个数为1时，输出结果中的number不加s，若个数为其他数时应加s

例如合法数字个数为3时

The average of 3 numbers is Y

合法数字为1时

The average of 3 number is Y

方法一：

#include<bits/stdc++.h>

using namespace std;
int main()
{
    string t;
    int n,i,num,d,j,flag,numd;
    double sum=0;
    cin>>n;
    num=0;  // 有效数字的个数
    for(i=1; i<=n; i++)
    {
        cin>>t;  //输入字符串
        numd=0; //记录小数点的个数
        flag=1; //记录有效数字的正负
        d=t.length();
        //判断字符串是否含有非法字符
        for(j=0; j<t.length(); j++)
        {
            if(j==0&&t[j]=='-')
            {
                flag=-1;
                continue;
            }
            else if(t[j]=='.')
            {
                d=j; //记录小数点的位置
                numd++;
            }
            else if(t[j]<'0'||t[j]>'9')
            {
                break;
            }
        }
        //若字符串首为小数点 则不合法
        if(t[0]=='.')
        {
            cout<<"ERROR: "<<t<<" is not a legal number"<<endl;
            continue;
        }
        //若存在非法字符，或者小数点个数大于1
        if(j!=t.length()||numd>=2)
        {
            cout<<"ERROR: "<<t<<" is not a legal number"<<endl;
            continue;
        }
        //计算小数位个数
        int xiao = t.length()-1-d;
        //如果小数位大于2则不合法
        if(xiao>2)
        {
            cout<<"ERROR: "<<t<<" is not a legal number"<<endl;
        }
        else
        {
            double ss = 0;
            for(j=0; j<t.length(); j++)
            {
                if(t[j]!='.'&&t[j]!='-')
                {
                    ss=ss*10+t[j]-'0';
                }
            }
            if(d!=t.length())
            {
                ss=ss*flag/pow(10,xiao); //转为小数
            }
            else
            {
                ss=ss*flag;
            }
            if(ss>=-1000&&ss<=1000)
            {
                sum+=ss;
                num++; //合法数字加一
            }
            else
            {
                cout<<"ERROR: "<<t<<" is not a legal number"<<endl;
            }
        }
    }
    if(num==0)
    {
        printf("The average of 0 numbers is Undefined\n");
    }
    else if(num==1)
    {
        printf("The average of %d number is %.2lf\n",num,sum/num);
    }
    else
    {
        printf("The average of %d numbers is %.2lf\n",num,sum/num);
    }
    return 0;
}

方法二：

使用下列两个函数可以快速解决这道题

sscanf() – 从一个字符串中读进与指定格式相符的数据

例如sscanf(a, "%lf", &temp);

它的作用是将字符串a中的符合浮点数格式的部分提取出来放到字符串temp中

例如a为1.23.14

前面的1.23符合浮点数格式。而后面的.14由于前面已经出现过小数点了，因此被判定为不符合浮点数格式，因此.14被丢弃

最终temp="1.12"
sprintf() – 字符串格式化命令，主要功能是把格式化的数据写入某个字符串中

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int main() {
    int n, cnt = 0;
    char a[50], b[50];
    double temp = 0.0, 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;
}