Operator

Problem:

In order to welcome Christmas, the students of the Informatics Interest Group held a grand party under the leadership of the tutors. The first content of the party was to play games: guess the number. The teacher gave each student a card with a number (this number is non-negative and less than 255). Each number is different from each other. The teacher made the following rules of the game: In the first round, each student squared the numbers of his number and then added them to get a group of numbers. The students with numbers in this group were eliminated; In the second round, the rest of the students cube the numbers they numbered to get a new set of numbers. The students whose numbers appear in this set of numbers will be eliminated; In the third round, the remaining students will add the numbers four times to get a new set of numbers. The students whose numbers appear in this set of numbers will be eliminated, and so on. After n rounds, the remaining students will get a special Christmas gift. The number on the card is the lucky number of 2009. (Assuming that the class size does not exceed 200)

Input format The first line is a positive integer n (n<8), indicating that there are n rounds of games The second line is the different numbers on the cards, separated by spaces. Output format Output a line for the remaining auspicious numbers, from small to large, with a space between each two numbers.

#include<bits/stdc++.h>
using namespace std;
class node{
	public:
		int value;
		int operator ^(int b){//Overload the "xor" operator to make it a power
			int tmp=b,tp=0,tp2=this->value;
			while(tp2){
				tp+=pow(tp2%10,tmp);
				tp2/=10;
			}
			return tp;
		}
};
node a[205];
int p,tab[205];
bool cmp(node a,node b){
	if(a.value<b.value)return 1;
	else return 0;
}
int main(){
	int n,t;
	cin>>n;
	while(cin>>a[p++].value);
	sort(a,a+p,cmp);
	for(int i=2;i<=n+1;i++){
		memset(tab,0,sizeof(tab));
		for(int j=0;j<p;j++)if(a[j].value!=-1)tab[j]=a[j]^i;
		for(int j=0;j<p;j++)
			for(int k=0;k<p;k++)if(tab[j]==a[k].value)a[k].value=-1;
	}
	for(int i=0;i<p;i++)if(a[i].value!=-1)cout<<a[i].value<<" ";
}