ZCMU--1367: Data Structure

Description

Give a set , Initially empty , Conduct N operations , There are three types of operations ：
1 Add an element to the collection , If it already exists in the collection , There is no need to add again
2 Remove an element from the collection , If the element does not exist in the collection , You do not need to delete
3 Determine the rank of the element in the set （ The youngest is the eldest ）, If the element does not exist, please output ："sorry"（ No double quotes ）

Input

Multiple sets of test data
Each group 1 That's ok ：1 It's an integer N, Indicates the number of operations .（2<=N<=10000）
The first 2 - （N+1） That's ok ： Each row 2 It's an integer k and s Corresponding to the mode of operation and the element to be operated （1<=k<=3,1<=s<=1000000）

Output

Corresponding to each operation 3, Give the corresponding results

Sample Input

8

3  1

2  1

1  1

1  1

1  2

3  2

2  2

3  2

5

1  1

1  100

1  1000000

2  99999

3   1000000

Sample Output

sorry

2

sorry

3

analysis ： Find out if the element exists , We can use arrays to quickly determine , But who is in the back row , violence 1~N The traversal must have timed out , So we have to use set To store , Then use the iterator to traverse , Initial settings c=1, It means who is in the row , Traverse ,c++, Until you find the element , Output c Is the rank of the element in the collection .

#include <bits/stdc++.h>
using namespace std;
int a[1000005];		// Used to directly determine whether an element exists  
int main()
{
	int n,z,p,c;
	while(~scanf("%d",&n)){
		set<int>st;	
		set<int>::iterator it;	// iterator , Prepare for the next ranking  
		memset(a,0,sizeof(a));	// initialization 0 
		while(n--){
			c=1;	// Who is the oldest  
			scanf("%d%d",&z,&p);
			if(z==1){		// Instructions 1 
				if(a[p]==0) st.insert(p),a[p]=1;// If the element does not exist , Deposit in set, also a[p] Turn into 1 
			}else if(z==2){		// Instructions 2 
				if(a[p]==1){	// If it exists, delete it  
					st.erase(st.find(p));	// Delete  
					a[p]=0;	//a[p] Set as 0, The element no longer exists  
				}
			}else if(z==3){
				if(a[p]==1){	// There is  
        			for (it=st.begin();it!=st.end();it++){	// Iterator traversal  
                	if(*it==p){		//*it Indicated value 
                		printf("%d\n",c);	// Found the element , Just output the ranking  
                		break;	
					}
					c++;	// Did not find , ranking ++ 
      			  }
				}else printf("sorry\n");	//a[p]=0, The element does not exist , Direct output sorry 
			}
		}
		st.clear();
	}
	return 0;
}