PAT_A1108#Finding Average

Source:

Description:

Input Specification:

Output Specification:

Sample Input 1:

Sample Output 1:

Sample Input 2:

Sample Output 2:

Keys:

• 字符串处理
• string（C++ STL）

Attention:

• 小数先相加再乘基数，整数先乘基数再相加
• str.c_str(）函数
• s.size()是unsigned，if（s.size()-pos-1 > 2）这个操作，实际上是无符号整数比较大小，真值的负数会大于正数

Code:

 /*

 Data: 2019-06-16 13:51:57

 Problem: PAT_A1108#Finding Average

 AC: 24:26

 题目大意：

 计算n个数的平均值，合法数范围在[-1000,1000]，且不超过两位小数

 */

 #include<cstdio>

 #include<string>

 #include<iostream>

 using namespace std;

 bool isLegal(string s, double &num)

 {

     int sign=;

     if(s[] == '-')

     {

         sign=-;

         s.erase(,);

     }

     int pos=s.size();

     for(int i=; i<s.size(); i++)

     {

         if(s[i] == '.')

         {

             if(pos==s.size())

                 pos = i;

             else

                 return false;

         }

         else if(s[i]<'' || s[i]>'')

             return false;

     }

     int len = s.size()-pos-;

     if(len > )

         return false;

     num=;

     double upper=,lower=;

     for(int i=; i<pos; i++)

     {

         upper *= ;

         upper += (s[i]-'');

     }

     for(int i=s.size()-; i>pos; i--)

     {

         lower += (s[i]-'');

         lower *= 0.1;

     }

     num = upper + lower;

     if(num > )

         return false;

     num *= sign;

     return true;

 }

 int main()

 {

 #ifdef  ONLINE_JUDGE

 #else

     freopen("Test.txt", "r", stdin);

 #endif // ONLINE_JUDGE

     int n,cnt=;

     string str;

     double sum=,num;

     scanf("%d", &n);

     for(int i=; i<n; i++)

     {

         cin >> str;

         if(isLegal(str,num))

         {

             cnt++;

             sum += num;

         }

         else

             printf("ERROR: %s is not a legal number\n", str.c_str());

     }

     if(cnt==)

         printf("The average of 0 numbers is Undefined");

     else if(cnt==)

         printf("The average of 1 number is %.2f", sum);

     else

         printf("The average of %d numbers is %.2f", cnt,sum/cnt);

     return ;

 }

Update：

 /*

 Data: 2019-08-17 20:23:41

 Problem: PAT_A1108#Finding Average

 AC: 24:26

 题目大意：

 计算n个数的平均值，合法数范围在[-1000,1000]，且不超过两位小数

 */

 #include<cstdio>

 #include<string>

 #include<iostream>

 using namespace std;

 const int Er=1e4;

 double ToF(string s)

 {

     int cnt=,acr=;

     string x="";

     for(int i=; i<s.size(); i++)

     {

         x += s[i];

         if(s[i]=='-' && i==)

             continue;

         if(cnt)

             acr++;

         if(s[i] == '.')

         {

             cnt++;

             if(cnt==)

                 return Er;

         }

         else if(s[i]<'' || s[i]>'')

             return Er;

     }

     if(acr>)

         return Er;

     else

         return atof(x.c_str());

 }

 int main()

 {

 #ifdef ONLINE_JUDGE

 #else

     freopen("Test.txt", "r", stdin);

 #endif // ONLINE_JUDGE

     int n,cnt=;

     double x=,sum=;

     string s;

     scanf("%d", &n);

     for(int i=; i<n; i++)

     {

         cin >> s;

         x = ToF(s);

         if(x>1e3 || x<-1e3)

             printf("ERROR: %s is not a legal number\n", s.c_str());

         else

         {

             sum += x;

             cnt++;

         }

     }

     if(cnt==)

         printf("The average of 0 numbers is Undefined\n");

     else if(cnt==)

         printf("The average of 1 number is %.2f\n", sum);

     else

         printf("The average of %d numbers is %.2f\n", cnt,sum/cnt);

     return ;

 }