PAT甲级 1041. Be Unique (20)

题目：

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, 10⁴]. 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 (<=10⁵) 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

思路：

之前做的思路是按顺序查找是否有unique值，有的输出，无则输出None。但结果证明其会有超时，即使把cin改成scanf，测试点4还是会超时。之后在输入时就利用数组统计次数，但查找时仍需要按照顺序，程序才通过。

代码：

#include<iostream>#include<vector>#include<fstream>using namespace std;int main(){int N;cin >> N;vector<int> squeue(N);int data[10001] = { 0 };//输入数据，并统计相关次数for (int i = 0; i < N; ++i){scanf("%d",&squeue[i]);data[squeue[i]]++;}//按顺序查找是否有unique的值bool flag = 0;for (int i = 0; i < N; ++i){if (data[squeue[i]] == 1){cout << squeue[i] << endl;flag = 1;break;}}//如果没有unique值if (!flag)cout << "None" << endl;system("pause");return 0;}