HDU 4802

本周作业的ProblemF。字符串类处理题。
传送门：http://acm.hdu.edu.cn/showproblem.php?pid=4802
GPA

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3596 Accepted Submission(s): 2164

Problem Description
In college, a student may take several courses. for each course i, he earns a certain credit (ci), and a mark ranging from A to F, which is comparable to a score (si), 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
2.33
0.00
4.00

Hint

For the first test case:
GPA =(3.0 * 2 + 1.0 * 3 + 0.0 * 1 + 4.0 * 3)/(2 + 3 + 1 + 3) = 2.33

For the second test case: because credit in GPA computation is 0(P/N in additional treatment), so his/her GPA is “0.00”.

题目没有什么坑的点，只要读准题意，然后写出来就行。
有些时候呢，这种题目也不要盲目的用strcmp(), 找到用于区分的特点，就可以写出来了。
AC代码如下：

#include <stdio.h>#include <string.h>#define maxn 1005struct cou{    int c;    char mark[2];} cou[maxn]; int main(){    int n;    while(scanf("%d", &n) != EOF && n)    {        double gpa = 0;        int sumc = 0;        double sc = 0;        for(int i = 0; i < n; i++)        {            scanf("%d %s", &cou[i].c, cou[i].mark);            if(cou[i].mark[0] == 'P' || cou[i].mark[0] == 'N')    continue;            sumc += cou[i].c;            if(cou[i].mark[1] == '\0')            {                switch(cou[i].mark[0])                {                    case 'A':                        sc += cou[i].c * 4.0;                        break;                    case 'B':                        sc += cou[i].c * 3.0;                        break;                    case 'C':                        sc += cou[i].c * 2.0;                        break;                    case 'D':                        sc += cou[i].c * 1.3;                        break;                    default:                        break;                }            }            else if(cou[i].mark[1] == '-')            {                switch(cou[i].mark[0])                {                    case 'A':                        sc += cou[i].c * 3.7;                        break;                    case 'B':                        sc += cou[i].c * 2.7;                        break;                    case 'C':                        sc += cou[i].c * 1.7;                        break;                    case 'D':                        sc += cou[i].c * 1.0;                        break;                    default:                        break;                }            }            else if(cou[i].mark[1] == '+')            {                switch(cou[i].mark[0])                {                    case 'B':                        sc += cou[i].c * 3.3;                        break;                    case 'C':                        sc += cou[i].c * 2.3;                        break;                    default:                        break;                }            }            gpa = sc / sumc;        }        printf("%.2f\n", gpa);    }    return 0;}