hdoj 4802 GPA 【水题】

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最新推荐文章于 2020-01-18 14:58:49 发布

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本文介绍了一个GPA计算程序，该程序能够根据学生的课程成绩和学分计算出加权平均绩点（GPA）。对于评分为P或N的课程，程序会自动忽略这些课程的成绩，确保计算出准确的GPA。

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GPA

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2405 Accepted Submission(s): 1457

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

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”.

题意：按照公式统计GPA，注意忽略那些评分为P或N的课程。当总学分为0时输出0。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (50000000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
double getscore(char *op)
{
    if(op[0] == 'A')
    {
        if(op[1] == '-')
            return 3.7;
        else
            return 4.0;
    }
    if(op[0] == 'B')
    {
        if(op[1] == '+')
            return 3.3;
        else if(op[1] == '-')
            return 2.7;
        else
            return 3.0;
    }
    if(op[0] == 'C')
    {
        if(op[1] == '+')
            return 2.3;
        else if(op[1] == '-')
            return 1.7;
        else
            return 2.0;
    }
    if(op[0] == 'D')
    {
        if(op[1] == '-')
            return 1.0;
        else
            return 1.3;
    }
    return 0;
}
int main()
{
    int n;
    while(Ri(n) != EOF)
    {
        double ans = 0;
        double Credit = 0;
        for(int i = 0; i < n; i++)
        {
            double a; char str[4];
            Rf(a); Rs(str);
            if(str[0] != 'N' && str[0] != 'P')
            {
                ans += a * getscore(str);
                Credit += a;
            }
        }
        if(Credit == 0)
            printf("0.00\n");
        else
            printf("%.2lf\n", ans / Credit);
    }
    return 0;
}