"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场（重现） HDU 5704

Luck Competition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11    Accepted Submission(s): 6

Problem Description

Participants of the Luck Competition choose a non-negative integer no more than 100 in their mind. After choosing their number, letK be the average of all numbers, and M be the result of K×23. Then the lucky person is the one who choose the highest number no more than M. If there are several such people, the lucky person is chosen randomly.

If you are given a chance to know how many people are participating the competition and what their numbers are, calculate the highest number with the highest probability to win assuming that you're joining the competition.

 

Input

There are several test cases and the first line contains the number of test casesT(T≤10).

Each test case begins with an integer N(1<N≤100), denoting the number of participants. And next line contains N−1 numbers representing the numbers chosen by other participants.

 

Output

For each test case, output an integer which you have chosen and the probability of winning (round to two digits after the decimal point), seperated by space.

 

Sample Input

341 2 341 1 2420 30 40

 

Sample Output

1 0.500 1.0018 1.00

 

Source

"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场

 

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水题 瞎JB模拟

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;//void printf(int begin,int end,int mid)//{//    int i;//    if(begin>end)//        return;//    for(i=begin; i<=end; i++)//        if(b[i]==a[mid])//            break;//    printf(begin,i-1,mid+1);  //中序，先左，后跟，再右。此时i为跟节点，左子树必定在begin到i-1里面。//    printf(i+1,end,mid-begin+i+1);//中序，先左，后跟，再右。此时i为跟节点，右子树必定在i+1到end里面。//    cout<<a[mid];//    if(mid==1)//        cout<<endl;//    else//        cout<<" ";//}//struct node//{//    int id;//    char s[1000];//};//bool cmp(node a,node b)//{//    return a.id>b.id;//}//char s[1005][1005];//int dir[4][2]= {1,0,-1,0,0,1,0,-1};//char g[5]= {"girl"};//char c[5]= {"cat"};//int n,m;//int dfs1(int x,int y,int cnt)//{//    int i;//    if(x<0 ||y<0 ||x>=n ||y>=m)//        return 0;//    if(s[x][y]!=g[cnt])//        return 0;//    if(cnt==3)//        return 1;//    int res=0;//    for(i=0; i<4; i++)//    {//        int dx=x+dir[i][0];//        int dy=y+dir[i][1];//        res+=dfs1(dx,dy,cnt+1);//    }//    return res;//}//int dfs2(int x,int y,int cnt)//{//    int i;//    if(x<0 ||y<0 ||x>=n ||y>=m)//        return 0;//    if(s[x][y]!=c[cnt])//        return 0;//    if(cnt==2)//        return 1;//    int res=0;//    for(i=0; i<4; i++)//    {//        int dx=x+dir[i][0];//        int dy=y+dir[i][1];//        res+=dfs2(dx,dy,cnt+1);//    }//    return res;//}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int i;        int n;        cin>>n;        int ans[105];        memset(ans,0,sizeof(ans));        int sum=0;        for(i=0; i<n-1; i++)        {            int x;            cin>>x;            sum+=x;            ans[x]++;        }        for(i=100; i>=0; i--)            ans[i]+=ans[i+1];        int id;        int res=343434;        int eee;        for(i=0; i<=100; i++)        {            int op=i+sum;            op=op*2/(3*n);            if(i<=op &&(i+1>op || ans[i+1]==0))  //前面没人，i为最接近op的。            {                if(ans[i]-ans[i+1]<=res)                {                    id=i;                    res=ans[i]-ans[i+1];                    eee=ans[i];                }            }        }       // cout<<eee<<endl;     //   cout<<res<<endl;      //  cout<<1.0/(eee+1)<<endl;        printf("%d %.2lf\n",id,1.0/(res+1));    }}