题解 | #频繁元素#

#include <iostream>
#include <algorithm>

using namespace std;

int main() {
    int nums[20], n, count = 0, max = 0, m;
    
    if(cin >> n) {
        for(int i = 0; i < n; i++) {
            cin >> nums[i];
        }
        
        for(int i = 0; i < n; i++) {
            count = 0;
            for(int j = 0; j < n; j++) {
                if(nums[i] == nums[j]) {
                    count++;
                }
                
                 if(count > max) {
                    max = count;
                    m = nums[i];
                 }
            }
        }
    }
    
    printf("%d\n", m);
    
    return 0;
}