There is sequence 1, 12, 123, 1234, ..., 12345678910, .... Now you are given two integers A and B, you have to find the number of integers from A^th number to B^th (inclusive) number, which are divisible by 3.

For example, let A = 3. B = 5. So, the numbers in the sequence are, 123, 1234, 12345. And 123, 12345 are divisible by 3. So, the result is 2.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains two integers A and B (1 ≤ A ≤ B < 2³¹) in a line.

Output

For each case, print the case number and the total numbers in the sequence between A^th and B^th which are divisible by 3.

Sample

Sample Input	Sample Output
Click to copy 2 3 5 10 110	Click to copy Case 1: 2 Case 2: 67

LOJ 1136 - Division by 3

Problem

Straight from the statement: "There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Now you are given two integers A and B, you have to find the number of integers from Ath number to Bth (inclusive) number, which are divisible by 3.

For example, let A = 3. B = 5. So, the numbers in the sequence are, 123, 1234, 12345. And 123, 12345 are divisible by 3. So, the result is 2."

Looking a bit into the sequence you will find out that for a certain position A the number can be found by simply appending all natural numbers from 1 upto A one after another. So if A = 5: sequence is 12345, if A = 12: sequence is 123456789101112.

Now you will be given two numbers A and B. You have to find out how many numbers on that sequence are divisible by 3 on the position of A, A+1, ..., B (from position A to position B inclusive).

Constrains:

Test cases T (≤ 10000) and 1 ≤ A ≤ B < 2^31

Observation

By looking at the constrains, it is quite sure that bruteforce approach is going to gift us TLE. (If you calculate all the number from range A to B and try to find out who amoung them are divisible by 3 you will get time limit exceeded). So we need a very effecient approach.
Numbers can be quite big.

Understanding 1

Let get our hands dirty by finding out if numbers from the sequence are divisible by 3.

Position at the sequence	Number	Is divisible by 3
1	1	false
2	12	true
3	123	true
4	1234	false
5	12345	true
6	123456	true
7	1234567	false
8	12345678	true
9	123456789	true
10	12345678910	false

We can clearly see a repeatitive pattern. Here {1, 4, 7, 10...} positions are not divisible by 3 and rests are.

Curious minds will want to know "why?". Simple!

First, you may know that calcuting if a number is divisible by 3 is super easy. You just add all digits and see if the sum is divisible by 3 or not. So finding out if 123456 is divisible by 3 is equal to finding out if (1 + 2 + 3 + 4 + 5 + 6) = 21 is divisible by 3 or not.

Now focus on the table. Each time we construct new position we just add the position-number to the previous value of the sequence, right? Let's follow an example. We have found out the 5th position in this sequence is 12345 which is divisible by 3 (12345 % 3 = 0; 12345 modulo 3 is zero. Also in other words (1 + 2 + 3 + 4 + 5) % 3 = 0). Now we construct 6th position in the sequence by just appending 6 to the last of the 5th number (123456). To calcuate if new number is divisible by 3 we can check that (1 + 2 + 3 + 4 + 5 + 6) % 3 is equal to 0 or not. Now after messing with paranthesis, you can see that (1 + 2 + 3 + 4 + 5 + 6) % 3 = ( (1 + 2 + 3 + 4 + 5) + 6) % 3. In other words for every n*, seq[n]* % 3 = (seq[n - 1] + n) % 3 = (seq[n - 1] % 3 + n % 3) % 3 (From modular arithmatic).

Now look at this table: | Position at the sequence | Number | Number % 3 | Previous_number % 3 | Position % 3 | (Previous_number % 3 + Position % 3) % 3 | |--------------------------|-------------|------------|---------------------|--------------|------------------------------------------| | 1 | 1 | 1 | - | 1 | Default value to 1 | | 2 | 12 | 0 | 1 | 2 | ( 1 % 3 + 2 % 3) % 3 = 0 | | 3 | 123 | 0 | 0 | 0 | ( 0 % 3 + 0 % 3) % 3 = 0 | | 4 | 1234 | 1 | 0 | 1 | ( 0 % 3 + 1 % 3) % 3 = 1 | | 5 | 12345 | 0 | 1 | 2 | ( 1 % 3 + 2 % 3) % 3 = 0 | | 6 | 123456 | 0 | 0 | 0 | ( 0 % 3 + 0 % 3) % 3 = 0 | | 7 | 1234567 | 1 | 0 | 1 | ( 0 % 3 + 1 % 3) % 3 = 1 | | 8 | 12345678 | 0 | 1 | 2 | ( 1 % 3 + 2 % 3) % 3 = 0 | | 9 | 123456789 | 0 | 0 | 0 | ( 0 % 3 + 0 % 3) % 3 = 0 | | 10 | 12345678910 | 1 | 0 | 1 | ( 0 % 3 + 1 % 3) % 3 = 1 |

Does the pattern make sense now?

Understanding 2

Previous undestanding is not enough to give you an AC, right? Now Let's look at the table a bit differently. Now we will add cumulative sum too and replace true/false with one/zero.

Position at the sequence	Number	Is divisible by 3	Cumulative sum over divisibility
1	1	0	0
2	12	1	1
3	123	1	2
4	1234	0	2
5	12345	1	3
6	123456	1	4
7	1234567	0	4
8	12345678	1	5
9	123456789	1	6
10	12345678910	0	6

Here cumulative sum for position n simply says "Between position 1 to n how many numbers are divisible by 3". Now can you tell how many numbers are divisible by 3 from position 4 to 7? We simply take the cumulative sum of position 7 and substract cumilative sum of position 3 (Because cumulative sum for position 7 says how many number are divisible by 3 from position 1 to 7, and we just substract how many numbers where divisible by 3 before the 4rth position).

Understanding 3

Now we know that cumulative sum can be the key to success. We just have to find it smartly. By looking at the last table we can get that {1, 4, 7, 10, ...} are not divisible by 3. And all this positions are also 1 modulo 3 (position % 3 = 1). So if we get a position which is 1 modulo 3 it will not be divisible, right? (If position % 3 = 1 then not divisible by 3.)

Now, finding out how many numbers are divisible by 3 from 1 to n is quite equal to finding how many numbers are not divisible by 3 and substract the answer from n. (numbers_divisible_by_3_from_1_to_n = n - numbers_not_divisible_by_3). Now our problem has become smaller. Here we can see that for every three consecutive position starting from 1, first position is not divisible by 3. So to calculate how many numbers are not divisible by 3 in range of position 1 to n...

If n % 3 = 0 then exactly (n / 3) integers are not divisible by 3.
If n % 3 = 1 then exactly floor(n / 3) + 1 integers are not divisible by 3.
If n % 3 = 2 then exactly floor(n / 3) + 1 integers are not divisible by 3. So now we will get how many numbers are divisible by 3 by simply substracting the result found above from the number n.

c++ function to calculate this is going to be (Beware of n = 1 case also)...

long long numbers_divisible_by_3_from_1_to_n (int n){
  if(n == 0) return 0;  // Caution: we will also call numbers_divisible_by_3_from_1_to_n(A - 1). If A = 1 then the argument becomes -1 which can result in     
                        // calculation.
  int subs;
  if(n % 3 == 0) {
    subs = n / 3;
  } else { // both n % 3 == 1 and n % 3 == 2 shows the same result
    subs = (n / 3) + 1;
  }
  return n - subs;
}

We solve this problem in O(1) right now. So as we will be given A and B and we need to find how many numbers from these range are divisible by 3 we simply do this.

long long result = numbers_divisible_by_3_from_1_to_n (B) - numbers_divisible_by_3_from_1_to_n (A - 1);

So overall complexity of the solution is O(1). Super handy!

C++

#include<bits/stdc++.h>
using namespace std;

int testcase;
long long A, B;

long long numbers_divisible_by_3_from_1_to_n (int n){
  if(n == 0) return 0;  // Caution: we will also call numbers_divisible_by_3_from_1_to_n(A - 1). If A = 1 then the argument becomes -1 which can result in     
                        // calculation.
  int subs;
  if(n % 3 == 0) {
    subs = n / 3;
  } else { // both n % 3 == 1 and n % 3 == 2 shows the same result
    subs = (n / 3) + 1;
  }
  return n - subs;
}

int main(){
  scanf("%d", &testcase);
  for(int test = 1; test <= testcase; test++){
    scanf("%lld %lld", &A, &B);
    long long result = numbers_divisible_by_3_from_1_to_n (B) - numbers_divisible_by_3_from_1_to_n (A - 1);
    printf("Case %d: %lld\n", test, result);
  }
}

Happy Coding!

Written by: Rahat Hossain

Tutorial source