PAT1041 Be Unique （20 分）

最新推荐文章于 2024-07-14 02:48:45 发布

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博客围绕火星独特的彩票规则展开，规则是从[1,104]选号，首个选到唯一数字者获胜。输入包含正整数N及N个投注数字，输出获胜数字，无则输出None。解析给出找首个唯一数字的方法，即统计各数字出现次数，再遍历输入数字判断。

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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

解析

找第一个unique的数字。最简单的方法：开一个arr[ $10^4+1$ ]的数字。把每个数字出现的次数+1。
最后就挨个遍历输入的数字：看出现的次数是不是1.

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
using namespace std;
int main()
{
	int N;
	scanf("%d", &N);
	vector<int> data(N,0);
	for (int i = 0; i < N; i++) 
		scanf("%d", &data[i]);
	map<int, int> Map;
	for (auto x : data)
		Map[x]++;
	vector<int> Unique;
	for (auto x : Map) {
		if (x.second == 1)
			Unique.push_back(x.first);
	}
	if (Unique.empty())
		printf("None");
	else {
		for (auto& x : Unique)
			x = find(data.cbegin(), data.cend(), x) - data.cbegin();
		printf("%d", data[*min_element(Unique.cbegin(), Unique.cend())]);
	}
}