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  • HDU-4968 Improving the GPA

    http://acm.hdu.edu.cn/showproblem.php?pid=4968

    这题是很简单的,我有思路,但是我确自己在比赛的时候没做,现在我按照我自己的思路写,一下子就a了,当时太不自信了,总是依靠同学去写。

    Improving the GPA

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 332    Accepted Submission(s): 268


    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
    
    
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    struct node
    {
        int low,hight;
        double v;
    }a[10];
    void init()
    {
        a[0].low=60;
        a[0].hight=69;
        a[0].v=2.0;
        a[1].low=70;
        a[1].hight=74;
        a[1].v=2.5;
        a[2].low=75;
        a[2].hight=79;
        a[2].v=3.0;
        a[3].low=80;
        a[3].hight=84;
        a[3].v=3.5;
        a[4].low=85;
        a[4].hight=100;
        a[4].v=4.0;
    }
    int main()
    {
        int t,avg,n,sum;
        int i,j,r,k,q;
        scanf("%d",&t);
        int s[10];
        double max,min,cnt;
        while(t--)
        {
             init();
              min=10;
              max=0;
             scanf("%d%d",&avg,&n);
            sum=avg*n;
            for(i=0;i<5;i++)
               s[i]=sum/a[i].low;
            for(i=0;i<=s[0];i++)
            {
                for(j=0;j<=s[1];j++)
                {
                    for(r=0;r<=s[2];r++)
                    {
                        for(k=0;k<=s[3];k++)
                        {
                            for(q=0;q<=s[4];q++)
                            {
                               if(i*a[0].low+j*a[1].low+r*a[2].low+k*a[3].low+q*a[4].low<=sum&&i+j+r+k+q==n)
                                {
                                     if(i*a[0].hight+j*a[1].hight+r*a[2].hight+k*a[3].hight+q*a[4].hight>=sum)
                                    {
                                      cnt=(i*a[0].v+j*a[1].v+r*a[2].v+k*a[3].v+q*a[4].v)/n;
                                        if(cnt>max)
                                            max=cnt;
                                        if(cnt<min)
                                           min=cnt;
                                    }
                                }
                            }
                        }
                    }
                }
            }
           printf("%.4lf %.4lf
    ",min,max);
        }
         return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/cancangood/p/3926378.html
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