HDU 1029 Ignatius and the Princess IV（快排orDP）

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本文介绍了一种算法问题，即从一组奇数个整数中找到出现次数超过一半的特殊整数，并提供了两种解决方案：一种使用快速排序，另一种采用线性时间复杂度的方法。

题目链接：http://hdu.hustoj.com/showproblem.php?pid=1029

题目大意:给定奇数n个数，找出出现次数大于等于（n+1）/2次的数

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
 

题目思路：

1.直接快排输出a[(n+1)/2];

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;
int a[N];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        printf("%d\n",a[(n+1)/2]);
    }
    return 0;
}

2.假定一个数为特殊数，若当前数与特殊数相同则cnt++，若不相同则cnt--，如果这时cnt<0，用当前数替代特殊的数。不管怎么样，由于特殊数次数大于(n+1)/2次，最终保留的数一定是特殊数。复杂度（O（n））

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int n;
    while(~scanf("%d",&n))
    {
        int num=0,ans,x;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(num==0)
            {
                num++;
                ans=x;
            }
            else
            {
                if(ans==x)
                    num++;
                else
                    num--;
            }
        }
        printf("%d\n",ans);

    }
    return 0;
}