hash学习pat1041 Be Unique

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博客围绕PAT考试中的一道算法题展开，题目是火星彩票以独特数字获胜，给定一组投注数字，首个投注唯一数字者获胜。输入包含投注人数和具体投注数字，输出获胜数字，若无则输出None，并给出了示例输入和输出。

1041 Be Unique (20 分)

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10
4
]. The first one who bets on a unique number wins. For example, if there are 7 people betting on { 5 31 5 88 67 88 17 }, then the second one who bets on 31 wins.

Input Specification:
Each input file contains one test case. Each case contains a line which begins with a positive integer N (≤10
5
) and then followed by N bets. The numbers are separated by a space.

Output Specification:
For each test case, print the winning number in a line. If there is no winner, print None instead.

Sample Input 1:
7 5 31 5 88 67 88 17
Sample Output 1:
31
Sample Input 2:
5 888 666 666 888 888
Sample Output 2:
None

#include <iostream>
#include<string>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
//Be Unique
//lottery 彩票抽奖 
/*
hash类的习题：
一般情况下首先去录入hash表，然后记录输入的数据
然后根据hash表来处理录入的数据 
题目大意：
 唯一的数字获胜
  
*/
bool cmp(int a,int b)
{
	return a>b;
}
using namespace std;
int main() {
     int n;
     int hash[10010]={0};
	 cin>>n; 
	 int *input=(int *)malloc(sizeof(int)*n);
  for(int i=0;i<n;i++)
  {
    cin>>input[i];
  	hash[input[i]]++;
  }
  bool flag=true;
  for(int i=0;i<n;i++)
  {
  	if(hash[input[i]]==1)
  	{
  		cout<<input[i];
  		flag=false;
  		break;
	  }
  }
  if(flag)
  {
  	printf("None");
  }
	system("pause");
	
}