HDU 5268 ZYB loves Score——BestCoder Round #44(模拟)

ZYB loves Score

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

24 012 020 0103 017 129 057 084 0

 

Sample Output

Case #1: 5722Case #2: 5412

 

Source

BestCoder Round #44

 

/************************************************************************/

附上该题对应的中文题

ZYB loves Score

 

 

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

问题描述

有一天$ZYB$参加了一场$BestCoder$，这场比赛一共有$4$道题，分数分别为$1000$,$1500$,$2000$,$2500$。一道题目如果在第$x$分钟解决，那么你只能得到这道题原来分数的$(250-x)/250\ast 100\%$由于原分数都是250的倍数，所以分数肯定是整数当一道题错误提交一次后，这道题的分数会额外降$50$分比如1000分的题你在5分钟时解决，然后你错误提交了一次，分数就是980-50=930为了防止分数过低，一道题的分数不会低于原来分数的$40\%$$ZYB$是个高手，他四道题在最后都通过了给出他四道题的过题时间和错误提交次数，求他最后的得分

输入描述

一共$T$($T\le 20$)组数据，对于每组数据：一共四行，每行一个非负整数$A$，$B$，表示这一题在$A$分钟获得Accept，错误提交了$B$次第$x$行表示第$x$题$0\le A\le 105$，$0\le B\le 100$

输出描述

每组数据输出一行Case #x: ans。x表示组数编号，从1开始。ans为所求值。

输入样例

24 012 020 0103 017 129 057 084 0

输出样例

Case #1: 5722Case #2: 5412

/****************************************************/

出题人的解题思路：

本题考察了选手的模拟能力，直接按照题目意思计算即可

本题题意很简单，就是按照题目的要求，给你一个选手做出4道题的时间及错误次数，求一下该选手参加一场BestCoder的最后得分。

4道题满分分别为1000、1500、2000、2500，假设该题满分为k分，那么若你做出该题的时间为t，错误的次数为c，则最终该题得分为

按照该公式分别算出4道题的得分，再求和即可

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<math.h>#include<vector>#include<map>#include<set>#include<cmath>#include<string>#include<algorithm>#include<iostream>#define exp 1e-10using namespace std;const int N = 50;const int inf = 1000000000;int fun(int t,int k,int c){    int n=2*c/5;    c=c/250*(250-t)-k*50;    return c>n?c:n;}int main(){    int t,i,ans,a,b,j=1;    scanf("%d",&t);    while(t--)    {        ans=0;        for(i=0;i<4;i++)        {            scanf("%d%d",&a,&b);            ans+=fun(a,b,1000+i*500);        }        printf("Case #%d: %d\n",j++,ans);    }    return 0;}

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