1041. Be Unique (20)【水题】——PAT (Advanced Level) Practise

最新推荐文章于 2017-08-03 20:51:15 发布

原创最新推荐文章于 2017-08-03 20:51:15 发布 · 1.6k 阅读

0 ·

CC 4.0 BY-SA版权

本文为博主原创文章，转载请标明出处 http://blog.youkuaiyun.com/xianyun2009

文章标签：

#pat #1041

浙大PAT 同时被 2 个专栏收录

106 篇文章

订阅专栏

浙大Pat

106 篇文章

订阅专栏

博客围绕PAT 1041题展开，介绍了题目信息，包括时间、内存、代码长度限制，以及输入输出规范和示例。解题思路是记录每个数出现次数，首个出现一次的数即为所求，最后给出了AC代码。

题目信息

1041. Be Unique (20)

时间限制100 ms
内存限制65536 kB
代码长度限制16000 B

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 (<=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:
31
Sample Input 2:
5 888 666 666 888 888
Sample Output 2:
None

解题思路

记录每个数出现次数，第一个出现一次的为所求

AC代码

#include <cstdio>
#include <map>
#include <vector>
using namespace std;
int main()
{
    int n, t;
    vector<int> v;
    map<int, int> mp;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i){
        scanf("%d", &t);
        v.push_back(t);
        mp[t]++;
    }
    bool flag = false;
    for (int i = 0; i < v.size(); ++i){
        if (mp[v[i]] == 1){
            printf("%d\n", v[i]);
            flag = true;
            break;
        }
    }
    if (!flag){
        printf("None\n");
    }
    return 0;
}