HDU4968Improving the GPA(分组背包)
Improving the GPA
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 283 Accepted Submission(s): 232
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.
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 3.2000HintIn 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 = 2.6667 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
题意：给一个n个人的平均数AVG，问计算出最小的GPA和最大的GPA。
解题：把总分数算出，就用分组背包进行DP，每个分数组合满足恰好的情况。
#include<stdio.h>
#define inf 99999999
int 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);
}
}