HDU 5268 ZYB loves Score (BestCoder Round#44)

题目链接：ZYB loves Score

题面：

ZYB loves Score

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 395    Accepted Submission(s): 232

Problem Description

One day,ZYB participated in the BestCoder Contest 

There are four problems. Their scores are 1000,1500,2000,2500 

According to the rules of BestCoder,If you solve one problem at  x  minutes,
You will get (250-x)/250 ∗ 100 %  of the original scores.

Obviously the final score must be an integer,becasue the original scores are multiple of 250

And if you make  x  wrong submissions，the score of this problem you get will be reduced by 50 ∗ x 

For example, if you solved the first problem in 5 minutes and you make one wrong submisson, the score of this problem is 980-50=930 

To prevent very low scores,the lowest score of one problem is  40%  of its original score

 

Input

Multiple test cases, the first line contains an integer T(no more than 20), indicating the number of cases. Each test case contains four lines

For i-th line of each test case there are two integers  A , B  which means you solved the i-th problem in A minutes and you have made B wrong submissons.

0≤A≤105 ， 0≤B≤100

 

Output

For each case, the output should occupies exactly one line. The output format is Case #x: ans, here x is the data number begins at 1.

 

Sample Input

   
   
   
   

    
    
    
    2
4 0
12 0
20 0
103 0
17 1
29 0
57 0
84 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 5722
Case #2: 5412
   
   
   
   

 

Source

BestCoder Round #44

 

解题：套下公式即可，注意分数不能低于40%。

代码：

#include <iostream>
#include <cmath> 
#define e  2.718281828459
using namespace std;
int max(double a,int b)
{
	if(a>b)return (int)a;
	else return b;
}
int main()
{
	int t,total,timee,wrong;
	cin>>t;
	for(int i=1;i<=t;i++)
	{
		total=0;
		for(int j=1000;j<=2500;j+=500)
		{
			cin>>timee>>wrong;
			total+=max((250.0-timee)/250*j-wrong*50,j*2/5);
		}
		cout<<"Case #"<<i<<": "<<total<<endl;
	}
   return 0; 
}