PAT 1041. Be Unique (20)(判断第一个不重复出现的数字是哪个)

最新推荐文章于 2020-05-04 17:07:48 发布

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最新推荐文章于 2020-05-04 17:07:48 发布

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博客介绍了1041. Be Unique (20)题目，包括时间、内存、代码长度限制等信息。该题是火星彩票规则相关，要求找出首个下注唯一数字者的数字，给出输入输出规范及示例，最后提及有AC代码。

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1041. Be Unique (20)

时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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

AC代码

#include<iostream>
#include<vector>
#include<string.h>
using namespace std;
int p[10001];
int main(int argc, char *argv[])
{
    int n;
    cin >> n;
    vector<int> a(n);
    //fill(p,p+10001,0);
    memset(p,0,sizeof(p));
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        p[a[i]]++;
    }
    //找出第一个只出现一次的，如果没有输出none
    int i;
    for (i = 0; i < n; i++) {
        if (p[a[i]] == 1) {
            break;
        }
    }
    if (i == n && p[a[n-1]] >1) {
        cout << "None" <<endl;
    }else {
        cout << a[i] << endl;
    }
    return 0;
}