hdu 5038 水题 但是题意坑

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

就是求个众数  这个范围小 所以一个数组存是否存在的状态就行了

但是这句话真恶心  If not all the value are the same but the frequencies of them are the same, there is no mode.

其实应该是这个意思：

当频率最高的有多个的时候，
如果 所有的grade出现的频率都是相等的，那么是没有mode的

否则按照升序

当然如果频率最高的有一个，还是有mode的

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;const int MAXN = 1e6+200;int g[MAXN];int a[MAXN],n,vis[MAXN];int cnt[MAXN];int out[MAXN];//map<int,int>cnt;int main(){    //IN("hdu5038.txt");    int ncase,n;    scanf("%d",&ncase);    for(int ic=1;ic<=ncase;ic++)    {        CL(cnt,0);        CL(vis,0);        scanf("%d",&n);        int mmax=0;//,mm=0;        for(int i=0;i<n;i++)        {             scanf("%d",&a[i]);             g[i]=10000 - (100-a[i])*(100-a[i]);             cnt[g[i]]++;             //vis[a[i]]=1;             mmax=max(mmax,cnt[g[i]]);            // mm=max(mm,g[i]);        }        int flag=0;        int cc=0;        for(int i=0;i<n;i++)        {            if(mmax == cnt[g[i]] && !vis[g[i]])            {                out[cc++]=g[i];                vis[g[i]]=1;            }            if(mmax != cnt[g[i]])            {                flag=1;            }        }        printf("Case #%d:\n",ic);        if(flag==0 && cc>1)puts("Bad Mushroom");        else        {            sort(out,out+cc);            printf("%d",out[0]);            int last=out[0];            for(int i=1;i<cc;i++)            {                   if(out[i]!=last)                   {                       last=out[i];                       printf(" %d",out[i]);                   }            }            putchar('\n');        }    }    return 0;}