2014北京網路賽1007||hdu5038 模擬

life4711發表於2014-09-23

http://acm.hdu.edu.cn/showproblem.php?pid=5038

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





這道題有必要說一下題目大意了，比賽時我們理解錯題意了==



題目大意：給你n個0~200之間的數，每個數對應有一個函式值，若所有的函式值出現的頻率都相等而自變數不相等時是Bad，否則輸出頻率最高的函式值，若有多個升序輸出。


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

int sum,num,cot[10006];
int T,n;

int main()
{
    int tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int num=0,sum=0;
        int maxx=0,minn=10005;
        memset(cot,0,sizeof(cot));
        for(int i=0;i<n;i++)
        {
            int w;
            scanf("%d",&w);
            int s=10000-(100-w)*(100-w);
            if(s>maxx) maxx=s;
            if(s<minn) minn=s;
            if(!cot[s])
                num++;
            cot[s]++;
            if(cot[s]>sum) sum=cot[s];
        }
        //printf("%d**%d\n",sum,num);
        printf("Case #%d:\n",++tt);
        if(sum*num==n&&num>1)
            printf("Bad Mushroom\n");
        else
        {
            int p=0;
            for(int i=minn;i<=maxx;i++)
            {
                if(cot[i]==sum)
                {
                    if(p)printf(" ");
                    printf("%d",i);
                    p=1;
                }
            }
            printf("\n");
        }
    }
    return 0;
}



            




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