HDU-4968 Improving the GPA （DP）

最新推荐文章于 2018-04-26 20:23:08 发布

原创最新推荐文章于 2018-04-26 20:23:08 发布 · 413 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#DP #hdu

HDU 同时被 2 个专栏收录

115 篇文章

订阅专栏

77 篇文章

订阅专栏

本文探讨了如何通过改进平均分数的计算方法，将分数转换为4分制等级点，并计算该等级点的最小和最大值。文章详细介绍了状态转移方程用于计算不同课程组合下的最小和最大GPA值，提供了具体的输入输出示例和实现代码。

Improving the GPA

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

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

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

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

了解到题意后很容易就能想到dp[i][j]表示前i门课程得分为j时点GPA总和

状态转移方程为：

mxdp[i][j]=max(mxdp[i][j],mxdp[i-1][j-k]+gpa);
mndp[i][j]=min(mndp[i][j],mndp[i-1][j-k]+gpa);

其中k表示第i门课程的成绩，取值范围为[60,100]，且mxdp[i-1][j-k]与mndp[i-1][j-k]为更新过的点

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int ave,n,sum,T;
int mndp[11][1001],mxdp[11][1001],gpa;

int getGpa(int sco) {//为了判断方便，都乘以10
    if(sco<70)
        return 20;
    if(sco<75)
        return 25;
    if(sco<80)
        return 30;
    if(sco<85)
        return 35;
    return 40;
}

int main() {
    scanf("%d",&T);
    while(T-->0) {
        scanf("%d%d",&ave,&n);
        sum=ave*n;
        memset(mxdp,-1,sizeof(mxdp));
        memset(mndp,0x3f,sizeof(mxdp));
        mxdp[0][0]=mndp[0][0]=0;
        for(int i=1;i<=n;++i) {
            for(int k=60;k<=100;++k) {
                gpa=getGpa(k);
                for(int j=k;j<=sum;++j) {
                    //if(mxdp[i-1][j-k]!=-1) {//mxdp和mndp同时更新，所以mxdp更新则mndp也必定更新，只有更新过的dp值才有效
                        mxdp[i][j]=max(mxdp[i][j],mxdp[i-1][j-k]+gpa);
                        mndp[i][j]=min(mndp[i][j],mndp[i-1][j-k]+gpa);
                    //}
                }
            }
        }
        printf("%.4lf %.4lf\n",0.1*mndp[n][sum]/n,0.1*mxdp[n][sum]/n);
    }
    return 0;
}