HDU 1029

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 14591    Accepted Submission(s): 5934

Problem 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

51 3 2 3 3111 1 1 1 1 5 5 5 5 5 571 1 1 1 1 1 1

 

Sample Output

351

 

题目大意： 计算出数组中个数大于 N+1 / 2 的数 

解题方法一 ： 可以认为是种位数， 因此只要将其进行排序就可以了；

#include<iostream>#include<algorithm>using namespace std;int a[1000000];int cmp(const void *a, const  void *b){    return *(int*)a - *(int*)b;}int main(){    int i, n;    while(cin>>n)    {        for(i = 0 ; i < n; i++)        {            cin>>a[i];        }        qsort(a, n, sizeof(int),cmp);        cout<<a[(n-1)/2]<<endl;    }}

编程之美中给来另外一种高效率的解法 ：   

编程之美  寻找发帖水王 

                即找到 ID大于一半的人

#include<iostream>using namespace  std;int a[1000000];int main (){    int n , i;    while(cin>>n)    {    for(i = 0 ; i < n; i++)    {        cin>>a[i];    }    int candate ;    int iTimes = 0;    for(i = 0; i < n; i++)    {//最后输出的candate 就会是结果        if(iTimes == 0)        {//从句收开始进行遍历            candate = a[i] ;            iTimes = 1;        }            else if(candate == a[i])            {//找到了一个相同的数字                iTimes++;            }            else            {//两个数字不相同                iTimes-- ;            }    }    cout<<candate<<endl;    }    return 0 ;}