HDU4802-GPA

GPA

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 814    Accepted Submission(s): 530

Problem Description

In college, a student may take several courses. for each course i, he earns a certain credit (c _i), and a mark ranging from A to F, which is comparable to a score (s

_i), according to the following conversion table

The GPA is the weighted average score of all courses one student may take, if we treat the credit as the weight. In other words,

An additional treatment is taken for special cases. Some courses are based on “Pass/Not pass” policy, where stude nts earn a mark “P” for “Pass” and a mark “N” for “Not pass”. Such courses are not supposed to be considered in computation. These special courses must be ignored for computing the correct GPA.

Specially, if a student’s credit in GPA computation is 0, his/her GPA will be “0.00”.

 

Input

There are several test cases, please process till EOF.
Each test case starts with a line containing one integer N (1 <= N <= 1000), the number of courses. Then follows N lines, each consisting the credit and the mark of one course. Credit is a positive integer and less than 10.

 

Output

For each test case, print the GPA (rounded to two decimal places) as the answer.

 

Sample Input

5 2 B 3 D- 2 P 1 F 3 A 2 2 P 2 N 6 4 A 3 A 3 A 4 A 3 A 3 A

 

Sample Output

Hint

 

Source

​​2013ACM/ICPC亚洲区南京站现场赛——题目重现 ​​

 

 

 

//注意浮点数和0的比较还有字符串的处理就OK了

AC代码：

#include<stdio.h>
#include<string.h>
char score[5];
double Fun(char score[])
{
 double x;
 if(score[0]>'D'&&score[0]!='F')
  return -1.0;
 else
 {
        if(strlen(score)==1)
  {
   switch(score[0])
   {
    case 'A':x=4.0;break;
    case 'B':x=3.0;break;
    case 'C':x=2.0;break;
    case 'D':x=1.3;break;
    case 'F':x=0;break;
   }
  }
  else
  {
   switch(score[0])
   {
    case 'A':x=3.7;break;
    case 'B':{
        switch(score[1])
        {
         case '+':x=3.3;break;
         case '-':x=2.7;break;
        }
       }break;
    case 'C':{
        switch(score[1])
        {
         case '+':x=2.3;break;
         case '-':x=1.7;break;
        }
       }break;
    case 'D':x=1.0;break;
   }
  }
  return x;
 }
}
int main()
{
 int i,n;
 double sum1,sum2,m;
 while(scanf("%d",&n)!=EOF)
 {
  sum1=0;sum2=0;
  for(i=0;i<n;i++)
  {
   memset(score,0,sizeof(score));
   scanf("%lf %s",&m,score);
   if(Fun(score)>-1.0)
   {
       sum1+=(m*Fun(score));
       sum2+=m;
   }
  }
  if(!(sum2>-0.000001&&sum2< 0.000001) )
      printf("%.2lf\n",sum1/sum2);
  else
   printf("0.00\n");
 }
 return 0;
}