Spoj Problem ACMCEG2C - Pick the candies (C++)

  The problem link may be found here.

      Explanation:

• Use Deque to keep track of elements of the variety of candies.
• If i is greater that allowed variety of Candies pop the front element of the queue
• Based of condition only keep the maximum element, pop everything thats smaller than the ith element

 1  #include<bits/stdc++.h>

 2  using namespace std;

 3  void sliding(vector<int>&v,int k){

 4  deque<int>q;

 5  vector<int>res;

 6  for(int i=0;i<v.size();i++){

 7      while(!q.empty() && i-q.front()>=k){

 8          q.pop_front();

 9      }

10      while(!q.empty() && v[q.back()]< v[i]){

11          q.pop_back();

12      }

13      q.push_back(i);

14      if(i>=k-1)res.push_back(v[q.front()]);

15  }

16  for(int i=0;i<res.size();i++){

17          cout<<res[i]<<" ";

18      }

19  }

20  int main(){

21  int t,n,k,m;

22 

23  scanf("%d",&t);

24  while(t--){

25      cin>>n>>k;

26       vector<int>v;

27       vector<int>ans;

28      for(int i=0;i<n;i++){

29          cin>>m;

30          v.push_back(m);

31      }

32      sliding(v,k);

33 

34  cout<<endl;

35  }

36 

37 

38  return 0;}

UVA Problem 12207 - That is Your Queue (C++)

The problem link can be found here. It has been solved using deque data structure.
Code:

 1  #include<bits/stdc++.h>

 2  using namespace std;

 3  int main(){

 4      int cases =0;

 5  int p,c;

 6 

 7  while(cin>>p>>c,p!=0,c!=0){

 8          deque<int>dq;

 9      if(p<c){

10          for(int i=1;i<=p;i++){

11            dq.push_back(i);

12          }

13      }

14      else{

15          for(int i=1;i<=c;i++){

16            dq.push_back(i);

17          }

18      }

19      cases++;

20      printf("Case %d:\n",cases);

21      while(c--){

22          string com;

23          cin>>com;

24          if(com == "N"){

25              int e = dq.front();

26              cout<<e<<endl;

27              dq.pop_front();

28              dq.push_back(e);

29          }

30          else{

31              int num;

32              cin>>num;

33              for(deque<int>::iterator it = dq.begin();it!=dq.end();it++){

34                  if(*it == num){

35                  dq.erase(it);

36                  break;

37                  }

38 

39              }

40              dq.push_front(num);

41          }

42      }

43  }

44 

45 

46  return 0;}