HDU4968Improving the GPA(分組背包)

標籤：dp   分組背包   

Improving the GPATime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 283    Accepted Submission(s): 232


Problem DescriptionXueba: 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.

InputThe 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).
OutputFor 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

475 175 275 375 10

Sample Output

3.0000 3.00002.7500 3.00002.6667 3.16672.4000 3.2000HintIn 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 = 2.6667Scores 79 78 68 GPA = (3.0 + 3.0 + 2.0) / 3 = 2.6667Scores 84 74 67 GPA = (3.5 + 2.5 + 2.0) / 3 = 2.6667Scores 100 64 61 GPA = (4.0 + 2.0 + 2.0) / 3 = 2.6667 

AuthorSYSU
Source2014 Multi-University Training Contest 9題意：給一個n個人的平均數AVG，問計算出最小的GPA和最大的GPA。解題：把總分數算出，就用分組背包進行DP，每個分數組合滿足恰好的情況。

#include<stdio.h>#define inf 99999999int main(){    double dp_max[11][1005],dp_min[11][1005];    int t,n,AVG,sum;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&AVG,&n);        sum=AVG*n;        for(int j=0;j<=n;j++)        for(int i=0;i<=sum;i++)        dp_max[j][i]=-1,dp_min[j][i]=inf;        dp_min[0][0]=dp_max[0][0]=0;        for(int i=1;i<=n;i++)        for(int s=sum;s>=60;s--)        for(int j=60;j<=100&&j<=s;j++)        {             if(j<70)            {                if(dp_max[i][s]<dp_max[i-1][s-j]+2.0&&dp_max[i-1][s-j]!=-1)                    dp_max[i][s]=dp_max[i-1][s-j]+2.0;                if(dp_min[i][s]>dp_min[i-1][s-j]+2.0)                    dp_min[i][s]=dp_min[i-1][s-j]+2.0;            }            else if(j<75)            {                if(dp_max[i][s]<dp_max[i-1][s-j]+2.5&&dp_max[i-1][s-j]!=-1)                    dp_max[i][s]=dp_max[i-1][s-j]+2.5;                if(dp_min[i][s]>dp_min[i-1][s-j]+2.5)                    dp_min[i][s]=dp_min[i-1][s-j]+2.5;            }            else if(j<80)            {                if(dp_max[i][s]<dp_max[i-1][s-j]+3.0&&dp_max[i-1][s-j]!=-1)                    dp_max[i][s]=dp_max[i-1][s-j]+3.0;                if(dp_min[i][s]>dp_min[i-1][s-j]+3.0)                    dp_min[i][s]=dp_min[i-1][s-j]+3.0;            }            else if(j<85)            {               if(dp_max[i][s]<dp_max[i-1][s-j]+3.5&&dp_max[i-1][s-j]!=-1)                    dp_max[i][s]=dp_max[i-1][s-j]+3.5;                if(dp_min[i][s]>dp_min[i-1][s-j]+3.5)                    dp_min[i][s]=dp_min[i-1][s-j]+3.5;            }            else            {                if(dp_max[i][s]<dp_max[i-1][s-j]+4.0&&dp_max[i-1][s-j]!=-1)                    dp_max[i][s]=dp_max[i-1][s-j]+4.0;                if(dp_min[i][s]>dp_min[i-1][s-j]+4.0)                    dp_min[i][s]=dp_min[i-1][s-j]+4.0;            }        }        printf("%.4lf %.4lf\n",dp_min[n][sum]/n,dp_max[n][sum]/n);    }}