hdu 5268 ZYB loves Score 水题
ZYB loves Score
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=5268
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
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
HINT
题意
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
题解：
本题考察了选手的模拟能力，直接按照题目意思计算即可
代码：
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 2000001 #define mod 10007 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //************************************************************************************** double a[4],b[4],c[4]; int main() { //test; int t=read(); c[0]=1000,c[1]=1500,c[2]=2000,c[3]=2500; for(int cas=1;cas<=t;cas++) { for(int i=0;i<4;i++) cin>>a[i]>>b[i]; int ans=0; for(int i=0;i<4;i++) ans+=max((250-a[i])/250*c[i]-b[i]*50,(c[i]*0.4)); printf("Case #%d: %d\n",cas,ans); } }
