【PAT刷题甲级】1041.Be Unique

最新推荐文章于 2021-11-15 11:28:47 发布

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该博客介绍了一个火星上独特的彩票系统，其中获胜规则是选择一个未被其他人选择过的数字。程序通过读取输入的投注数列，统计每个数字出现的频率，输出第一个只出现一次的数字作为赢家。如果所有数字都重复，则输出None。提供的代码实现了这一逻辑，通过遍历投注并检查次数来找到获胜者。

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

Sample Input 2

5 888 666 666 888 888

Sample Output 2

None

题意

对输入的数字的出现次数进行统计，输出第一个只出现1次的数字。（散列）

代码

#include <iostream>
#include <vector>
using namespace std;
int bets[100010];
vector<int> v;
int main(){
	int n,k,flag=0;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&k);
		v.push_back(k);
		bets[k]++;
	}
	for(int i=0;i<v.size();i++){
		if(bets[v[i]]==1){
			printf("%d",v[i]);
			return 0;
		}	
	}
	printf("None");
	return 0;
}