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  • hdu 1084(水题)

    10分钟左右A了...

                What Is Your Grade?

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6195    Accepted Submission(s): 1867


    Problem Description
    “Point, point, life of student!”
    This is a ballad(歌谣)well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course.
    There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically(以此类推), you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50.
    Note, only 1 student will get the score 95 when 3 students have solved 4 problems.
    I wish you all can pass the exam! 
    Come on!
     
    Input
    Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T(consumed time). You can assume that all data are different when 0<p.
    A test case starting with a negative integer terminates the input and this test case should not to be processed.
     
    Output
    Output the scores of N students in N lines for each case, and there is a blank line after each case.
     
    Sample Input
    4 5 06:30:17 4 07:31:27 4 08:12:12 4 05:23:13 1 5 06:30:17 -1
     
    Sample Output
    100 90 90 95 100
     
    Author
    lcy
     
     
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    
    struct node
    {
        int key;
        int a,b,c;
        int id,cos;
    }g[110];
    
    int cnt[10];
    
    int cmp(node x,node y)
    {
        if(x.key!=y.key) return x.key>y.key;
        if(x.a!=y.a) return x.a<y.a;
        if(x.b!=y.b) return x.b<y.b;
        if(x.c!=y.c) return x.c<y.c;
    }
    
    int save[110];
    
    int main()
    {
        int n;
        while(scanf("%d",&n)&&n>0)
        {
            memset(cnt,0,sizeof(cnt));
            for(int i=0;i<n;i++)
            {
                scanf("%d %d:%d:%d",&g[i].key,&g[i].a,&g[i].b,&g[i].c);
                cnt[g[i].key]++;
                g[i].id=i;
            }
            sort(g,g+n,cmp);
            int tcnt=0;//
            int k=-1;
            for(int i=0;i<n;i++)
            {
                if(g[i].key!=k) 
                {
                    k=g[i].key;
                    tcnt=1;
                }
                else tcnt++;
                if(g[i].key==0) 
                {
                    save[g[i].id]=50;
                    continue;
                }
                if(g[i].key==5)
                {
                    save[g[i].id]=100;
                    continue;
                }
                int tmp=100-10*(5-g[i].key);
                if(tcnt <= cnt[g[i].key]/2) tmp+=5;
                save[g[i].id]=tmp;
            }
            for(int i=0;i<n;i++)
                printf("%d\n",save[i]);
            printf("\n");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/chenhuan001/p/2951180.html
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