怒刷DP之 HDU 1029-白红宇

怒刷DP之 HDU 1029

阅读量：6124 次

发布时间：2019-06-21

本文共 1928 字，大约阅读时间需要 6 分钟。

Ignatius and the Princess IV

Time Limit:1000MS Memory Limit:32767KB 64bit IO Format:%I64d & %I64u

Submit

Appoint description: jehad (2014-05-27)gscsdlz (2015-04-14)System Crawler (2015-09-05)

Description

"OK, you are not too bad, em... But you can never pass the next test." feng5166 says.

"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.

"But what is the characteristic of the special integer?" Ignatius asks.

"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.

Can you find the special integer for Ignatius?

Input

The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file.

Output

For each test case, you have to output only one line which contains the special number you have found.

Sample Input

5 1 3 2 3 3 11 1 1 1 1 1 5 5 5 5 5 5 7 1 1 1 1 1 1 1

Sample Output

3 5 1

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
            7 #include 
            
              8 #include 
             
               9 #include 
              
               10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  13 #include 
                  
                   14 using namespace std;15 16 int main(void)17 {18 int n,box,count,max;19 20 while(scanf("%d",&n) != EOF)21 {22 count = 0;23 while(n --)24 {25 scanf("%d",&box);26 27 if(!count)28 {29 max = box;30 count ++;31 }32 else33 if(max == box)34 count ++;35 else36 count --;37 }38 printf("%d\n",max);39 }40 41 return 0;42 }

转载于:https://www.cnblogs.com/xz816111/p/4789382.html

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