• "Accepted today?"[HDU1177]


    "Accepted today?"

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2832    Accepted Submission(s): 1290


    Problem Description
    Do you remember a sentence "Accepted today?" Yes, the sentence is mentioned frequently in lcy's course "ACM Programming"!
    The contest is still in progress this moment. How excited it is! You, smart programmer, must have AC some problems today. "Can I get copper medal, silver medal, or even golden medal?" Oh, ha-ha! You must be considering this question. And now, the last problem of this contest comes.
    Give you all submitting data in the contest, and tell you the number of golden medals, silver medals and copper medals; your task is to output someone's contest result.
    Easy? Of course! I t is the reason that I designed the problem.
    When you have completed this contest, please remember that sentence〃 Accepted today?〃兒
     

     

    Input
    Input contains multiple test cases. Each test case starts with five numbers N (4 =< N <= 130 -- the total number of attendees), G, S, C (1<=G<=S<=C<N --G, S, C denoting the number of golden medals, silver medals and copper medals respectively) and M (0<M<=N). The next N lines contain an integer P (1<=P<=8 --number of problems that have been solved by someone) and a time T(for example,"02:45:17", meaning 2 hours and 45 minutes and 17 seconds consumed according to contest rules) each. You can assume that all submit data are different.
    A test case starting with 0 0 0 0 0 terminates input and this test case should not to be processed.
     

     

    Output
    For each case, print a sentence in a line, and it must be one of these sentences:
    Accepted today? I've got a golden medal :)
    Accepted today? I've got a silver medal :)
    Accepted today? I've got a copper medal :)
    Accepted today? I've got an honor mentioned :)

    Note:
    You will get an honor mentioned if you can't get copper medal, silver medal or golden medal.
     

     

    Sample Input

    10 1 2 3 6
    2 02:45:17
    2 02:49:01
    2 03:17:58
    2 03:21:29
    4 07:55:48
    3 04:25:42
    3 06:57:39
    2 02:05:02
    2 02:16:45
    2 02:41:37
    0 0 0 0 0

     

    Sample Output
    Accepted today? I've got a silver medal :)
    #include <stdio.h>  
    #include <algorithm>  
    using namespace std;  
      
    struct AC  
    {  
        int num,h,m,s;  
        int no;  
        int k;//k=1金,2银,3铜,4,安慰奖  
    } a[200];  
      
    int cmp(AC x,AC y)  
    {  
        if(x.num!=y.num)  
        return x.num>y.num;  
        if(x.h!=y.h)  
        return x.h<y.h;  
        if(x.m!=y.m)  
        return x.m<y.m;  
        return x.s<y.s;  
    }  
      
    int main()  
    {  
        int n,g,s,c,m,i;  
        while(~scanf("%d%d%d%d%d",&n,&g,&s,&c,&m))  
        {  
            if(n+g+s+m+c == 0)  
                break;  
            for(i = 0; i<n; i++)  
            {  
                a[i].no = i+1;  
                scanf("%d %d:%d:%d",&a[i].num,&a[i].h,&a[i].m,&a[i].s);  
            }  
            sort(a,a+n,cmp);  
            //printf("====
    ");  
            for(i = 0; i<n; i++)  
            {  
                if(g)  
                {  
                    a[i].k = 1;  
                    g--;  
                }  
                else if(s)  
                {  
                    a[i].k = 2;  
                    s--;  
                }  
                else if(c)  
                {  
                    a[i].k = 3;  
                    c--;  
                }  
                else  
                {  
                    a[i].k = 4;  
                }  
            }  
            for(i = 0; i<n; i++)  
            {  
                if(a[i].no == m)  
                {  
                    if(a[i].k == 1)  
                        printf("Accepted today? I've got a golden medal :)
    ");  
                    else if(a[i].k == 2)  
                        printf("Accepted today? I've got a silver medal :)
    ");  
                    else if(a[i].k == 3)  
                        printf("Accepted today? I've got a copper medal :)
    ");  
                    else if(a[i].k == 4)  
                        printf("Accepted today? I've got an honor mentioned :)
    ");  
                    break;  
                }  
            }  
        }  
      
        return 0;  
    }  
    View Code

     

  • 相关阅读:
    人工智能,垂直领域
    西方企业该向中国式创新学什么?
    [Leetcode 6] ZigZag问题的一种新思路
    Unknown column &#39;*&#39; in &#39;field list&#39; 异常解决
    多媒体——图像文件大小的计算
    Lucky Number
    win7 64位系统下进入debug
    Chromium网页DOM Tree创建过程分析
    做好长期奋斗的准备
    IP寻址和子网运算
  • 原文地址:https://www.cnblogs.com/dramstadt/p/6086947.html
Copyright © 2020-2023  润新知