Ignatius and the Princess IV
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 22321 Accepted Submission(s): 9335
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
这是著名的“寻找多数元素”（“主元素问题”）。
我首先想到的是建一个times数组，记录每个数出现的次数，当每个数出现次数大于N+1/2的时候就可以输出了。
之后我又想可以对所有数排一次序，然后第N/2个数一定就是主元素。
最后我看了看discuss区，又写了如下的算法。
//
// main.cpp
// 基础DP1-B-Ignatius and the Princess IV
//
// Created by 袁子涵 on 15/10/22.
// Copyright ? 2015年 袁子涵. All rights reserved.
//#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string>long long int N;
int sum,num,out;int main(int argc,const char*argv[])
{while (scanf("%lld",&N)!=EOF) {sum=0;for (long long int i=0; i<N; i++) {scanf("%d",&num);if (sum==0) {sum++;out=num;}else{if (num!=out) {sum--;}elsesum++;}}printf("%d\n",out);}
}