模拟

就遇力扣到周赛原题了
如果不在当前编号下，就一直交换。时间复杂度$O(n)$

#include <iostream>

using namespace std;

const int N = 10010;

int n;
int a[N];

int main()
{
    cin >> n;
    for(int i = 1;i <= n;i ++ ) cin >> a[i];

    int cnt = 0;
    for(int i = 1;i <= n;i ++ )
    {
        while(a[i] != i){
            swap(a[a[i]],a[i]);
            cnt ++;
        }
    }

    cout << cnt << endl;
}