hdu---（5038）Grade（胡搞）

Grade

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 109    Accepted Submission(s): 63

Problem Description

Ted is a employee of Always Cook Mushroom (ACM). His boss Matt gives him a pack of mushrooms and ask him to grade each mushroom according to its weight. Suppose the weight of a mushroom is w, then it’s grade s is s = 10000 - (100 - w)^2 What’s more, Ted also has to report the mode of the grade of these mushrooms. The mode is the value that appears most often. Mode may not be unique. If not all the value are the same but the frequencies of them are the same, there is no mode.

Input

The first line of the input contains an integer T, denoting the number of testcases. Then T test cases follow. The first line of each test cases contains one integers N (1<=N<=10^6),denoting the number of the mushroom. The second line contains N integers, denoting the weight of each mushroom. The weight is greater than 0, and less than 200.

Output

For each test case, output 2 lines. The first line contains "Case #x:", where x is the case number (starting from 1) The second line contains the mode of the grade of the given mushrooms. If there exists multiple modes, output them in ascending order. If there exists no mode, output “Bad Mushroom”.

Sample Input

3 6 100 100 100 99 98 101 6 100 100 100 99 99 101 6 100 100 98 99 99 97

Sample Output

Case #1: 10000 Case #2: Bad Mushroom Case #3: 9999 10000

Source

2014 ACM/ICPC Asia Regional Beijing Online

代码： 此题的意思就是给你一些蘑菇，依据蘑菇的重量得到不同的品质，有品质得到价格，当然100的品质最好。求出现频率最好的品质，如果有多组，以升序输出。

 1 #include<cstdio>
 2 #include<cstring>
 3 int cnt[205];
 4 int main()
 5 {
 6   int w,cas,n,i,j;
 7   scanf("%d",&cas);
 8   for( i=1;i<=cas;i++)
 9   {
10       scanf("%d",&n);
11       memset(cnt,0,sizeof(cnt));
12       for( j=1;j<=n;j++)
13       {
14       scanf("%d",&w);
15       if(w>100) w=200-w;
16       cnt[w]++;  //统计出现的频率
17       }
18       int maxc=0;  //最大频率
19       for( j=0;j<=100;j++)
20        if(cnt[maxc]<cnt[j])
21            maxc=j;
22 
23     printf("Case #%d:\n",i);
24     for( j=0;j<=100;j++)
25       if(cnt[j]<cnt[maxc]&&cnt[j]>0) break;
26 
27       if(j>100){
28         if(cnt[maxc]==n)
29             printf("%d\n",10000-(100-maxc)*(100-maxc));
30         else
31             printf("Bad Mushroom\n");
32       }
33       else
34       {
35           for( j=0;j<=100;j++)
36             if(cnt[maxc]==cnt[j]){
37             printf("%d",10000-(100-j)*(100-j));
38             j++;
39             break;
40         }
41         for( ;j<=100;j++)
42             if(cnt[maxc]==cnt[j])
43             printf(" %d",10000-(100-j)*(100-j));
44             printf("\n");
45       }
46  }
47   return 0;
48 }