PAT 1041 Be Unique_pat1041 be unique-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_50679551/article/details/126974506

这道PAT 1041题目要求在一组彩票号码中找出第一个出现且仅出现一次的数字作为赢家。输入包含一个正整数N和N个彩票号码，需要输出获胜号码。如果所有号码都重复，则输出'No Winner'。解决方案可以通过记录每个号码出现次数的数组实现，按输入顺序检查唯一数字。

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1041 Be Unique

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

总结：这道题目还是比较简单的

思路：暴力做法，数组 a 存放一个数的出现次数，数组 b 存放题目给出数的顺序，在照数组 b 存的顺序查看那个数第一次出现且只出现了一次的，这个就是结果

#include <iostream>
using namespace std;

const int N=100010;
int a[N],b[N];
int n;

int main(){
    cin >> n;
    
    for(int i=0;i<n;i++){
        scanf("%d",&b[i]);
        a[b[i]]++;
    }
    
    for(int i=0;i<n;i++){
        if(a[b[i]]==1){
            printf("%d\n",b[i]);
            return 0;
        }
    }
    
    printf("None\n");
    
    return 0;
}

好好学习，天天向上！

我要考研！

2022.11.5(map会比直接使用hash耗时1.5倍左右吧)

//哈希表
#include <iostream>
using namespace std;

int ha[10010];
int main(){
    int n;
    scanf("%d",&n);
    int a[n]={0};
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        ha[a[i]]++;
    }
    for(int i=0;i<n;i++){
        if(ha[a[i]]==1){
            printf("%d",a[i]);
            return 0;
        }
    }
    printf("None\n");
    return 0;
}

//map的方法
#include <iostream>
#include <map>
using namespace std;

map<int,int> m;
int main(){
    int n;
    scanf("%d",&n);
    int a[n];
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        m[a[i]]++;
    }
    for(int i=0;i<n;i++){
        if(m[a[i]]==1){
            printf("%d\n",a[i]);
            return 0;
        }
    }
    printf("None\n");
    return 0;
}