ZOJ3705:Applications

Recently, the ACM/ICPC team of Marjar University decided to choose some new members from freshmen to take part in the ACM/ICPC competitions of the next season. As a traditional elite university in ACM/ICPC, there is no doubt that application forms will fill up the mailbox. To constitute some powerful teams, coaches of the ACM/ICPC team decided to use a system to score all applicants, the rules are described as below. Please note that the score of an applicant is measured by pts, which is short for "points".

1. Of course, the number of solved ACM/ICPC problems of a applicant is important. Proudly, Marjar University have a best online judge system called Marjar Online Judge System V2.0, and in short, MOJ. All solved problems in MOJ of a applicant will be scored under these rules:

• (1) The problems in a set, called MaoMao Selection, will be counted as 2.5 pts for a problem.
• (2) The problems from Old Surgeon Contest, will be counted as 1.5 pts for a problem.There is no problem in MaoMao Selection from Old Surgeon Contest.
• (3) Besides the problem from MaoMao Selection and Old Surgeon Contest, if the problem's id is a prime, then it will be counted as 1 pts.
• (4) If a solved problem doesn't meet above three condition, then it will be counted as 0.3 pts.

 

2. Competitions also show the strength of an applicant. Marjar University holds the ACM/ICPC competition of whole school once a year. To get some pts from the competition, an applicant should fulfill rules as below:

• The member of a team will be counted as 36 pts if the team won first prize in the competition.
• The member of a team will be counted as 27 pts if the team won second prize in the competition.
• The member of a team will be counted as 18 pts if the team won third prize in the competition.
• Otherwise, 0 pts will be counted.

 

3. We all know that some websites held problem solving contest regularly, such as JapanJam, ZacaiForces and so on. The registered member of JapanJam will have a rating after each contest held by it. Coaches thinks that the third highest rating in JapanJam of an applicant is good to show his/her ability, so the scoring formula is:

 

Pts = max(0, (r - 1200) / 100) * 1.5

 

Here r is the third highest rating in JapanJam of an applicant.

4. And most importantly - if the applicant is a girl, then the score will be added by 33 pts.

The system is so complicated that it becomes a huge problem for coaches when calculating the score of all applicants. Please help coaches to choose the best M applicants!

Input

There are multiple test cases.

The first line of input is an integer T (1 ≤ T ≤ 10), indicating the number of test cases.

For each test case, first line contains two integers N (1 ≤ N ≤ 500) - the number of applicants and M (1 ≤ M ≤ N) - the number of members coaches want to choose.

The following line contains an integer R followed by R (0 ≤ R ≤ 500) numbers, indicating the id of R problems in MaoMao Selection.

And then the following line contains an integer S (0 ≤ S ≤ 500) followed by S numbers, indicating the id of S problems from Old Surgeon Contest.

The following line contains an integer Q (0 ≤ Q ≤ 500) - There are Q teams took part in Marjar University's competition.

Following Q lines, each line contains a string - team name and one integer - prize the team get. More specifically, 1 means first prize, 2 means second prize, 3 means third prize, and 0 means no prize.

In the end of each test case, there are N parts. In each part, first line contains two strings - the applicant's name and his/her team name in Marjar University's competition, a char sex - M for male, F for female and two integers P (0 ≤ P ≤ 1000) - the number of problem the applicant solved, C (0 ≤ C ≤ 1000) - the number of competitions the applicant have taken part in JapanJam.

The following line contains P integers, indicating the id of the solved problems of this applicant.

And, the following line contains C integers, means the rating for C competitions the applicant have taken part in.

We promise:

• The problems' id in MaoMao Selection, Old Surgeon Contest and applicant's solving list are distinct, and all of them have 4 digits (such as 1010).
• All names don't contain spaces, and length of each name is less than 30.
• All ratings are non-negative integers and less than 3500.

 

Output

For each test case, output M lines, means that M applicants and their scores. Please output these informations by sorting scores in descending order. If two applicants have the same rating, then sort their names in alphabet order. The score should be rounded to 3 decimal points.

Sample Input

1
5 3
3 1001 1002 1003
4 1004 1005 1006 1007
3
MagicGirl!!! 3
Sister's_noise 2
NexusHD+NexusHD 1
Edward EKaDiYaKanWen M 5 3
1001 1003 1005 1007 1009
1800 1800 1800
FScarlet MagicGirl!!! F 3 5
1004 1005 1007
1300 1400 1500 1600 1700
A NexusHD+NexusHD M 0 0


B None F 0 0


IamMM Sister's_noise M 15 1
1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015
3000

Sample Output

FScarlet 60.000
IamMM 44.300
A 36.000

 

题意：这题题目很长，条件也非常多，导致我们比赛的时候少看了一个条件，一直纠结了很久

第一行给出一个数字，代表样例的个数

第二行给出两个数，代表人数和输出排名的前几个人

第三四行分别给出MaoMao Selection和Surgeon Contest的题目数量与题号

第五行给出拿到前三等奖的数目，接下来给出每个获奖的队伍和获得的排名

最后给出每个人的信息，包括人名，队名，性别，OJ里的做题数，参加比赛的数目

然后给出题号与比赛的得分

而每个人的得分计算是这样：

首先是做的题所得的分：1.在MaoMao Selection做的题得2.5分

                    2.在Surgeon Contest做的题得1.5分

                                              3.如果不在这两个地方做的题并且题号为奇数，得1分

                                               4.否则0.3分

如果所在的队伍得奖了：一等奖36分，尔等27，三等18

如果是女的：得到33分

如果参加过比赛，则取排第三的分数，计算公式Pts = max(0, (r - 1200) / 100) * 1.5为得分，如果小于3次比赛则不算

 

知道了题意之后，就是简单的模拟题了

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct TEAM
{
    char name[50];
    int rank;
} team[505];

struct MAN
{
    char name[30],team[30];
    char sex[3];
    double score;
    int pro,rush;
    int p[505],r[505];
} man[505];

int mao[505],sur[505],k_mao,k_sur;

int prime(int x)
{
    int i;
    for(i = 2; i*i<=x; i++)
        if(x%i==0)
            return 1;
}

double problem(int x)
{
    int i;
    for(i = 0; i<k_mao; i++)
    {
        if(x == mao[i])
            return 2.5;
    }
    for(i = 0; i<k_sur; i++)
    {
        if(x == sur[i])
            ;
        return 1.5;
    }
    if(prime(x))
        return 1;
    return 0.3;
}

int cmp_rank(TEAM x,TEAM y)
{
    return x.rank<y.rank;
}

double big(double a,double b)
{
    return a>b?a:b;
}

int cmp_score(MAN x,MAN y)
{
    return x.score>y.score;
}

int cmp_r(double a,double b)
{
    return a>b;
}

int main()
{
    int t,man_num,team_num,m,i,j,k,xx;
    double ss,r[1005];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&man_num,&m);
        scanf("%d",&k_mao);
        for(i = 0; i<k_mao; i++)
            scanf("%d",&mao[i]);
        scanf("%d",&k_sur);
        for(i = 0; i<k_sur; i++)
            scanf("%d",&sur[i]);
        scanf("%d",&team_num);
        for(i = 0; i<team_num; i++)
            scanf("%s%d",team[i].name,&team[i].rank);
        for(i = 0; i<man_num; i++)
        {
            man[i].score = 0;
            scanf("%s%s%s%d%d",man[i].name,man[i].team,man[i].sex,&man[i].pro,&man[i].rush);

            for(j = 0; j<man[i].pro; j++)
            {
                scanf("%d",&xx);
                man[i].score+=problem(xx);
            }
            sort(team,team+team_num,cmp_rank);
            for(j = 0; j<team_num; j++)
            {
                if(!strcmp(man[i].team,team[j].name))
                {
                    if(team[j].rank==1)
                        man[i].score+=36;
                    else if(team[j].rank==2)
                        man[i].score+=27;
                    else if(team[j].rank==3)
                        man[i].score+=18;
                }
            }
            memset(r,0,sizeof(r));
            for(j = 0; j<man[i].rush; j++)
            {
                scanf("%lf",&r[j]);
            }
            sort(r,r+man[i].rush,cmp_r);
            if(man[i].rush>=3)
                man[i].score+=big(0,((r[2]-1200.0)/100)*1.5);
            if(!strcmp(man[i].sex,"F"))
                man[i].score+=33;
        }
        sort(man,man+man_num,cmp_score);
        for(i = 0; i<m; i++)
        {
            printf("%s %.3lf\n",man[i].name,man[i].score);
        }
    }
    return 0;
}