PAT (Advanced Level) Practise 1041 Be Unique (20)

最新推荐文章于 2018-02-22 14:08:26 发布

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最新推荐文章于 2018-02-22 14:08:26 发布

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博客介绍了火星独特的彩票规则，即从[1, 104]选号，首个选到唯一数字者获胜。给出输入文件包含正整数N及N个投注数，要求输出获胜数字，无获胜者则输出“None”，实质是输出只出现一次的数。

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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:

Sample Input 2:

5 888 666 666 888 888

Sample Output 2:

None

题意：给你n个数，输出其中只出现了一次的数

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

const int N=200010;
const int INF=0x7FFFFFFF;

int a[100005];
int x[100005];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(x,0,sizeof x);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            x[a[i]]++;
        }
        int flag=0;
        for(int i=0;i<n;i++)
        {
            if(x[a[i]]==1)
            {
                flag=1;
                printf("%d\n",a[i]);
                break;
            }
        }
        if(!flag) printf("None\n");
    }
    return 0;
}