HDU 1029 Ignatius and the Princess IV(基础dp（思维）)

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本文介绍了一个算法问题，即从一组给定的整数中找出出现次数超过(n+1)/2的特殊整数，并提供了一种高效的解决方案。通过使用两个变量x和y来追踪可能的特殊整数及其出现频率。

[kuangbin带你飞]专题十二基础DP1

网址：https://cn.vjudge.net/contest/68966#problem/B

Ignatius and the Princess IV

HDU - 1029

"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

题意：给你n个数，让你求其中出现了大于等于（n+1）/2次的数。

没用dp，设置两个变量x，y。x存数，y存次数。每次输入一个数进行比较，如果与x相同，y++，如果不相同y--，若y为0，x变化，因为结果出现了（n+1）/2次，因此最后x一定为结果。

代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 2147483647


int main()
{
    int t;
    int a;
    int x,y;
    while(scanf("%d",&t) != EOF){
        y = 0;
        x = -MAXN-1;
        for(int i = 0;i < t;i++){
            scanf("%d",&a);
            if(a == x){
                y++;
            }
            else if(y == 0){
                x = a;
            }
            else{
                y--;
            }
        }
        printf("%d\n",x);
    }
    return 0;
}