You don't need to declare an iterator object just to check if the value exists. And switch looks more suitable here than lot of if-then...

int main() {
int iCount;
set<int> ss;
cin >> iCount;
for (int i=0; i<iCount; ++i){
    int type, query;
    cin >> type >> query;
    switch (type){
        case 1:
            ss.insert(query);
            break;
        case 2:
            ss.erase(query);
            break;
        case 3:
            cout << (ss.find(query) == ss.end() ? "No" : "Yes") << endl;
            break;
    }
}  
return 0;
}

wow... what a poor wording this problem has... it explains what i should do if Q=1 or 2?... do I need to study the given example to find out what the problem is all about??

Have the problem solved, but do not know why if using

if (*itr == x) cout << "Yes" << endl;

would fail test case #10 and #12. I think it should be the same as

if (itr != s.end()) cout << "Yes" << endl;

Any hints?

full code as follows:

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int q, n, i, j, y, x;
    cin >> q;
    set<int> s;
    for (i=0;i<q;i++) {
        cin >> y >> x;
        if (y == 1) {
            s.insert(x);
        }
        if (y ==2) {
            s.erase(x);
        }
        if (y == 3) {
            set<int>::iterator itr = s.find(x);
            if (itr != s.end()) cout << "Yes" << endl;
            //if (*itr == x) cout << "Yes" << endl;
            else cout << "No" << endl;
        }
    }
    return 0;
}

This works:

it=s.find(d);
if(it!=s.end())
    printf("Yes\n");
else
    printf("No\n");

But this gives wrong answer in 2 testcases:

it=s.find(d);
if( (*it)==d)
    printf("Yes\n");
else
    printf("No\n");

Can I know why?

TRy THIs its work fine.. /

include

include

include

int main()

{ long long n,x,y;

cin>>n;
set<long long> s;
for(long long i=0;i<n;i++)
    {cin>>x>>y;
    if(x==1)
        {s.insert(y);
         }
    else if(x==2)
    {s.erase(y);
         }else {set<long long>::iterator it=s.find(y);
         if(it==s.end()) // if we use *it==y,got error
         {cout<<"No\n";
         }else{cout<<"Yes\n";
         }
         }   
     }
 return 0;

}