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
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
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
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∗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≤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≤A≤105，0≤B≤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
2 4 0 12 0 20 0 103 0 17 1 29 0 57 0 84 0
输出样例
Case #1: 5722 Case #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-10
using 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;
}菜鸟成长记