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;}

SPOJ Problem ABSP1 Solution in (C++)

Welcome:

The question to the problem can be found here:


Explanation:
1.Notice the array is sorted already.
2. It can be solved with simple Math.
3. There are two logic only.
4. Multiply each element  i -th times(i-th times: position of element in the array)
5. Multiply each element n-1-i -th times.
6. Subtract 4 and 5.
7.Happy Coding !

Code:

#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;

long long ABS(long long a[],long long n){
    long long sum=0;
rep(i,n){
sum += (a[i]*i) - (a[i]*(n-1-i));
}
return sum;
}
int main(){
 //   freopen("input.txt","r",stdin);
 //freopen("output.txt","w",stdout);
int t;
long long arr[10000];
cin>>t;
while(t-->0){
long long n;
cin>>n;
rep(i,n){
    cin>>arr[i];
}
cout<<ABS(arr,n)<<endl;

}

return 0;}