Problem Description
In college, a student may take several courses. for each course i, he earns a certain credit (c), and a mark ranging from A to F, which is comparable to a score (s), according to the following conversion table
![HDOJ 4802 GPA-LMLPHP]( "HDOJ 4802 GPA-LMLPHP")
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,
![HDOJ 4802 GPA-LMLPHP]( "HDOJ 4802 GPA-LMLPHP")
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”.
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.
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
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
2.33 0.00 4.00
思路：输入数据后进行判断，如果是标准的GPA，则返回相应的数据，并用一个全局变量来表示是否全部都是非法的GPA，如果是则直接输出0即可
#include <iostream>
#include <string>
using namespace std;
int flag=;//全局变量，用来标识是否全是P或者N
float table(char *p)
{
//flag=0;
if(p[]=='P'||p[]=='N')
return 0.0;
else
{
if(p[]=='A'&&p[]!='-')
{flag++;return 4.0;}
else if(p[]=='A'&&p[]=='-')
{flag++;return 3.7;}
else if(p[]=='B'&&p[]=='+')
{flag++;return 3.3;}
else if(p[]=='B'&&p[]!='-'&&p[]!='+')
{flag++;return 3.0;}
else if(p[]=='B'&&p[]=='-')
{flag++;return 2.7;}
else if(p[]=='C'&&p[]=='+')
{flag++;return 2.3;}
else if(p[]=='C'&&p[]!='-'&&p[]!='+')
{flag++;return 2.0;}
else if(p[]=='C'&&p[]=='-')
{flag++;return 1.7;}
else if(p[]=='D'&&p[]!='-')
{flag++;return 1.3;}
else if(p[]=='D'&&p[]=='-')
{flag++;return 1.0;}
else {flag++;return 0.0;}
}
}
int main()
{
int n=;
int i=;
float GPA=0.0;
int c;
int ci=;
char s[]={};
while(scanf("%d",&n)!=EOF)
{
GPA=0.0;
ci=0.0;
flag=;
for(i=;i<n;i++)
{
cin>>c>>s;
GPA+=((float)c)*table(s);
if(!(s[]=='P'||s[]=='N'))
ci+=c;
}
if(flag==)
printf("0.00\n");
else
{
GPA=GPA/(float)ci;
printf("%.2f\n",GPA);
}
}
}