Problem:
Score : 300 points

Problem Statement

You are given two non-negative integers L and R.
We will choose two integers i and j such that L \leq i < j \leq R.
Find the minimum possible value of (i \times j) \mbox{ mod } 2019.

Constraints

All values in input are integers.

0 \leq L < R \leq 2 \times 10^9

Input

Input is given from Standard Input in the following format:

L R

Output

Print the minimum possible value of (i \times j) \mbox{ mod } 2019 when i and j are chosen under the given condition.

Sample Input 1

2020 2040

Sample Output 1

2

When (i, j) = (2020, 2021), (i \times j) \mbox{ mod } 2019 = 2.

Sample Input 2

4 5

Sample Output 2

20

We have only one choice: (i, j) = (4, 5).

Solution in PHP (5.6.30) (judge verdict: Accepted)

<?php
fscanf(STDIN,"%d %d",$L,$R);
$lastCalc = false;
if($R-$L>=2019){
  //$R-$Lが2019以上なら間に2019の倍数を含むので最小剰余は必ず0
  $result = 0;
}elseif($R*($R-1)<2019){
  //$L*$Rが2019以上でない場合、最小剰余は$L*($L+1)
  //$result = (($L%2019)*(($L+1)%2019))%2019;
  //一時的に0にし総当りループを抜けた後に計算させる
  $result = 0;
  $lastCalc = true;
}else{
  //1組目を初期値にする
  $result = (($L%2019)*(($L+1)%2019))%2019;
}
//総当り
for($i=$L;$i<=$R-1;$i++){
  if($result === 0){
    break;
  }
  for($j=$i+1;$j<=$R;$j++){
    if($result === 0){
      break;
    }
    //最小値更新
    if($result > (($i%2019)*($j%2019))%2019){
      $result = (($i%2019)*($j%2019))%2019;
    }
  }
}
//if文を1行で書くと{}を省略できる
if($lastCalc) $result = $L*($L+1);
echo $result;