HDU 1029Ignatius and the Princess IV

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 32991 Accepted Submission(s): 14186

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
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

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思路

1、n是奇数，(n+1)/2是n的一半以上，只要将n个数据排序，出现(n+1)/2次的整数必然会出现在中间位置。

2、因为N(1<=N<=999999) ，所以数组不能为局部变量，否则导致栈溢出，所以要定义为全局数组

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代码

#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;int a[1000000]={0};int main(){    int n;    while(cin>>n)    {        for(int i=0;i<n;i++)        {            cin>>a[i];        }        sort(a,a+n);//升序排序        cout<<a[(n+1)/2]<<endl;    }    return 0;}

总结

如果定义的局部变量数组大小过大，编译器会报错“栈溢出”。
数组声明在函数内部，属于局部变量，存放在了栈上，如果数组过大比如a[1000000]。那数组占用的内存大小为：1000000*4byte约等于4M。而栈的默认内存空间为1M左右，所以会导致内存溢出解决这个问题。
解决这个问题方法为：可以将数组声明在全局存储区或堆上即可
方法：申明为全局变量