| 23 | return 0; |
| 24 | } |
| 25 | void solve() |
| 26 | { |
| 27 | long long int noOfDigits, first, second, sum=0, remaining, arr[4], temp; |
| 28 | cin>>noOfDigits>>first>>second; |
| 29 | sum+=first+second; |
| 30 | if((first+second)%5==0) //logic explained above |
| 31 | { |
| 32 | cout<<"NO\n"; |
| 33 | } |
| 34 | else if((first+second)%2==0) |
| 35 | { |
| 36 | noOfDigits-=2; //as we have already added two digits sum in variable sum |
| 37 | sum+=(noOfDigits/4)*20; //as (2+4+6+8=20) |
| 38 | remaining=noOfDigits%4; |
| 39 | arr[0]=(first+second)%10; |
| 40 | arr[1]=(first+second+arr[0])%10; |
| 41 | arr[2]=(first+second+arr[0]+arr[1])%10; |
| 42 | arr[3]=(first+second+arr[0]+arr[1]+arr[2])%10; |
| 43 | for(int i=0;i<remaining;i++) |
| 44 | { |
| 45 | sum+=arr[i]; |
| 46 | } |
| 47 | if(sum%3==0) |
| 48 | cout<<"YES\n"; |
| 49 | else |
| 50 | cout<<"NO\n"; |
| 51 | } |
| 52 | else |
| 53 | { |
| 54 | temp=first+second; |
| 55 | first=temp; //considering sum of first two digits as first digit |
| 56 | second=temp%10; //third digit |
| 57 | sum+=second; |
| 58 | noOfDigits-=3; //as we have already added two digits sum in variable sum |
| 59 | sum+=(noOfDigits/4)*20;//as (2+4+6+8=20) |
| 60 | remaining=noOfDigits%4; |
| 61 | arr[0]=(first+second)%10; |
| 62 | arr[1]=(first+second+arr[0])%10; |
| 63 | arr[2]=(first+second+arr[0]+arr[1])%10; |
| 64 | arr[3]=(first+second+arr[0]+arr[1]+arr[2])%10; |
| 65 | for(int i=0;i<remaining;i++) |
| 66 | { |
| 67 | sum+=arr[i]; |
| 68 | } |
| 69 | if(sum%3==0) |
| 70 | cout<<"YES\n"; |
| 71 | else |
| 72 | cout<<"NO\n"; |
| 73 | } |
| 74 | } |