HDU 4968 Improving the GPA

Improving the GPA

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 652    Accepted Submission(s): 476

Problem Description

Xueba: Using the 4-Point Scale, my GPA is 4.0.

In fact, the AVERAGE SCORE of Xueba is calculated by the following formula:
AVERAGE SCORE = ∑(Wi * SCOREi) / ∑(Wi) 1<=i<=N
where SCOREi represents the scores of the ith course and Wi represents the credit of the corresponding course.

To simplify the problem, we assume that the credit of each course is
1. In this way, the AVERAGE SCORE is ∑(SCOREi) / N. In addition, SCOREi
are all integers between 60 and 100, and we guarantee that ∑(SCOREi) can
be divided by N.

In SYSU, the university usually uses the
AVERAGE SCORE as the standard to represent the students’ level. However,
when the students want to study further in foreign countries, other
universities will use the 4-Point Scale to represent the students’
level. There are 2 ways of transforming each score to 4-Point Scale.
Here is one of them.

The student’s average GPA in the 4-Point Scale is calculated as follows:GPA = ∑(GPAi) / N
So given one student’s AVERAGE SCORE and the number of the courses,
there are many different possible values in the 4-Point Scale. Please
calculate the minimum and maximum value of the GPA in the 4-Point Scale.

 

Input

The input begins with a line containing an integer T (1 < T <
500), which denotes the number of test cases. The next T lines each
contain two integers AVGSCORE, N (60 <= AVGSCORE <= 100, 1 <= N
<= 10).

 

Output

For each test case, you should display the minimum and maximum value
of the GPA in the 4-Point Scale in one line, accurate up to 4 decimal
places. There is a space between two values.

 

Sample Input

4

75 1

75 2

75 3

75 10

 

Sample Output

3.0000 3.0000

2.7500 3.0000

2.6667 3.1667

2.4000 3.2000

Hint

In the third case, there are many possible ways to calculate the minimum value of the GPA in the 4-Point Scale.
For example,

Scores 78 74 73 GPA = (3.0 + 2.5 + 2.5) / 3 = 2.6667

Scores 79 78 68 GPA = (3.0 + 3.0 + 2.0) / 3 = 2.6667

Scores 84 74 67 GPA = (3.5 + 2.5 + 2.0) / 3 = 2.6667

Scores 100 64 61 GPA = (4.0 + 2.0 + 2.0) / 3 = 2.6667

 

Author

SYSU

 

Source

2014 Multi-University Training Contest 9

 

 

 

解析：题中GPA分为5种档次，N <= 10，数据量比较小，可以直接枚举每种档次的GPA的课程数。对于每种可能方案，算出该方案下的最低总分和最高总分。如果输入的总分在此范围内，说明该方案是可行的。算出所有可行方案中的最小GPA和最大GPA即可。

 

 

 

 #include <cstdio>
 
 int T;
 int AVGSCORE,N;
 int sumscore;
 int max_score,min_score;
 double max_gpa,min_gpa;
 
 int main()
 {
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&AVGSCORE,&N);
         sumscore = AVGSCORE*N;
         max_gpa = -1.0;
         min_gpa = 0x7fffffff;
         for(int i = ; i <= N; ++i)
             for(int j = ; j<= N-i; ++j)
                 for(int k = ; k <= N-i-j; ++k)
                     for(int l = ; l <= N-i-j-k; ++l){
                         int m = N-i-j-k-l;
                         max_score = *i+*j+*k+*l+*m;
                         min_score = *i+*j+*k+*l+*m;
                         if(sumscore <= max_score && sumscore >= min_score){
                             double gpa = 4.0*i+3.5*j+3.0*k+2.5*l+2.0*m;
                             if(gpa>max_gpa)
                                 max_gpa = gpa;
                             if(gpa<min_gpa)
                                 min_gpa = gpa;
                         }
                     }
         max_gpa /= N;
         min_gpa /= N;
         printf("%.4f %.4f\n",min_gpa,max_gpa);
     }
     return ;
 }