1041. Be Unique (20)-PAT甲级真题（Hash散列）

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
题目大意：给n个数字，按照读入顺序，哪个数字是第一个在所有数字中只出现一次的数字。如果所有数字出现都超过了一次，则输出None
分析：建立一个数组，存储每个数字出现的次数，然后遍历一遍输入的顺序看是否有出现次数为1的数字

#include <cstdio>
using namespace std;
int a[100001], m[100000];
int main() {
    int n;
    scanf("%d", &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", a[i]);
            return 0;
        }
    }
    printf("None");
    return 0;
}

#include <cstdio>

using namespace std;

int a[100001], m[100000];

int main() {

int n;

scanf("%d", &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", a[i]);

return 0;

}

printf("None");

return 0;

}

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