PAT_A 1041. Be Unique (20)

PAT_A 1041. Be Unique (20)

时间限制 100 ms 内存限制 65536 kBBeing 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 17Sample Output 1:31Sample Input 2:5 888 666 666 888 888Sample Output 2:None

• 分析：

  • 这题要求，找出最先出现的那个数，且出现一次。时间限制100ms，数据最大10^5,写完程序，担心超时，好比t1039。索性没有太多要求，一次AC。
  • 方法：利用stl的map+vector完成，map用来完成重复数据计算，vector用来标识数据次序。最后按vector先后顺序，查找第一个map中value=1的key,即是。

• code:

#include<iostream>#include<cstdio>#include<map>#include<vector>using namespace std;map<int,int> people;vector<int> seq;int main(){    freopen("in","r",stdin);    int N,tmp;    cin>>N;    for(int i=0;i<N;i++)    {        cin>>tmp;        if(people[tmp]==0)          seq.push_back(tmp);        people[tmp]++;    }    int i=0;    int size=seq.size();    for(i=0;i<size;i++)    {        if(people[seq[i]]==1)        {          cout<<seq[i]<<endl;          break;        }    }    if(i==size)      cout<<"None"<<endl;    return 0;}

• AC