621A. Wet Shark and Odd and Even

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5. https://codeforces.com/contest/621/problem/A
6. A. Wet Shark and Odd and Even

    Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.

   Note, that if Wet Shark uses no integers from the n integers, the sum is an even integer 0.

   Input

   The first line of the input contains one integer, n (1 ≤ n ≤ 100 000). The next line contains n space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.

   Output

   Print the maximum possible even sum that can be obtained if we use some of the given integers.

   Examples

   input

   Copy

   3
   1 2 3

   output

   Copy

   6

   input

   Copy

   5
   999999999 999999999 999999999 999999999 999999999

   output

   Copy

   3999999996

   Note

   In the first sample, we can simply take all three integers for a total sum of 6.

   In the second sample Wet Shark should take any four out of five integers 999 999 999.
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11. #include<bits/stdc++.h>
12. using namespace std;
13. int main(){
14. long long n,od=0,ev=0,sum=0;
15. cin>>n;
16. long long a[n];
17. for(int i=0;i<n;i++){
18. cin>>a[i];
19. if(a[i]%2) od++;
20. else ev++;
21. }
22. sort(a,a+n);
23. reverse(a,a+n);
24. for(int i=0;i<n;i++){
25. if(a[i]%2==0) sum=sum+a[i];
26. }
27. if(od%2==0){
28. for(int i=0;i<n;i++)
29. if(a[i]%2==1) sum=sum+a[i];
30. }
31.  
32. else if(od%2==1){int c=0;
33. for(int i=0;i<n;i++)
34. if(a[i]%2==1)
35. {
36. if(c==od-1) break;
37. else
38. sum=sum+a[i];
39. c++;
40. }
41. }
42. cout<<sum;
43.  
44.  
45. return 0;
46. }