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Luck Competition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1051    Accepted Submission(s): 646

Problem Description Participants of the Luck Competition choose a non-negative integer no more than 100 in their mind. After choosing their number, let  K  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 cases  T(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 
  
  
   
   3
4
1 2 3
4
1 1 2
4
20 30 40
 
Sample Output 
  
  
   
   1 0.50
0 1.00
18 1.00
 
Source "巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场  
Recommend liuyiding
题意：给你 n - 1个数，你要选出一个数，使你选出的数在不大于这 n 个数的平均数 × 2/3的前提下尽量大； 
#include <bits/stdc++.h>
using namespace std;

int t, n, sum;
int a[1001];

int judge(int k){
    //printf("k = %d  %f\n", k, (double)(k+sum)/n*2/3);
    if((double)(k+sum)/n*2/3 < k) return 0;
    return 1;
}

int main(){
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        sum = 0;
        for(int i = 1; i < n; i++){
            scanf("%d", &a[i]);
            sum += a[i];
        }
        int l = 0, r = 100, ans;
        while(l <= r){
            int mid = (l+r)>>1;
            if(judge(mid)) ans = mid, l = mid+1;
            else r = mid-1;
        }
        int cnt = 1;
        for(int i = 1; i < n; i++){
            if(a[i] == ans) cnt++;
        }
        printf("%d %.2f\n", ans, 1.0/cnt);
    }
}




本文标签： hducompetitionLuck

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