独一无二的题解

#include<iostream>
using namespace std;
int n,d,dp[1005][105],a[1005];
int main(){
	cin>>n>>d;
	a[1]=-1;
	dp[1][0] = 1;//1的0级上司
	for(int i=2;i<=n;i++){
		cin>>a[i];
		dp[i][0] = i;
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=d;j++){
			if(dp[i][j-1]!=-1) dp[i][j] = a[dp[i][j-1]];
			else dp[i][j] = -1;
		}
	}
	for(int i=1;i<=n;i++){
		cout<<dp[i][d]<<endl;
	}
    return 0;
}