Grade

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 360    Accepted Submission(s): 203

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

题意：给你一些蘑菇，由他的重量求出价值

让你输出价值中出现次数最多的。如果有多个价值出现一样多，并且没有别的价值了，那么就输出Bad Mushroom。

否则输出出现频率最高的价值，按照升序输出。

因为价值最多是0-10000 所以开个数组记录下出现的次数就好了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
using namespace std;
int a[10010];
int cnt = 0, T, N;
int main(){
	scanf("%d", &T);
	while(T--){
		memset(a, 0, sizeof(a));
		cnt++; printf("Case #%d:\n", cnt);

		scanf("%d", &N);
		int temp, max = 0;
		int mark = N;
		while(N--){
			scanf("%d", &temp);
			a[10000-(100-temp)*(100-temp)]++;
		}

		for(int i = 0; i <= 10000; i++){
			if(a[i] > max)  max = a[i];
		}
		int cnt2 = 0;
		for(int i = 0; i <= 10000; i++){
			if(a[i] == max) cnt2++;
		}
		if(cnt2*max == mark && cnt2 > 1) printf("Bad Mushroom\n");
		else{
			bool flag2 = false;
			for(int i = 0; i <= 10000; i++){
				if(a[i] == max) {
					if(flag2 == false){
						printf("%d", i); flag2 = true;
					}
					else printf(" %d", i);
				}
			}
			printf("\n");
		}

	}

	return 0;
}