题解 | #频繁元素#

#include <bits/stdc++.h> using namespace std; int main() { int n; cin >> n; int a[100];

for( int i = 0 ; i < n ; i ++ )   
{
    cin >> a[i];
}

int s = 0 , ans = 0;
for( int i = 0 ; i < n ; i ++ )
{
    int cnt = 0;
    for( int j = 0 ; j < n ; j ++ ) 
    {
        if(a[i] == a[j]) cnt ++;
    }
    if(s < cnt)
    {
        s = cnt;
        ans = a[i];
    }
}
cout << ans;

}