Question

Lili was working on her mathematic homework, and suddenly, she found another classic problem.

The problem is: “Given three light bulbs X , Y , and Z. Bulb X light up every A seconds, bulbs Y light up

every B seconds, and bulb Z light up every C seconds. If three bulbs light up together for the 1-st

time at 0-th second, the 2-nd time this three bulbs will light up together at the same time will be at

K1-th second. Find K1!”

Lili got bored by this question. Lili challenge herself and she figured out a challenging problem. She

wants to find the sum of all Ki which satisfy L ≤ Ki ≤ R, where when at Ki-th seconds, the three bulbs

light up together for the i + 1-th time. Solve Lili’s challenge!

Format Input :

There are T testcases. Every testcase consists of five integers A , B , C , L, and R as described above.

Format Output:

Output T testcases with format “Case # X: ”, where X indicates the testcase number and then

followed by an integer indicates the answer to Lili’s challenge.

Constraints:

• 1 ≤ T ≤ 10

• 1 ≤ A, B, C ≤ 100

• 1 ≤ L ≤ R ≤ 109

Sample Input (standard input):

5

1 2 3 1 20

2 3 5 30 30

2 3 5 30 90

11 11 11 10 40

2 3 4 10 30

Sample Output (standard output):

Case #1: 36

Case #2: 30

Case #3: 180

Case #4: 66

Case #5: 36

NOTE: USE C LANGUAGE, DONT USE FUNCTION(RESULT,RETURN),VOID,RECURSIVE,

USE BASIC CODE AND CODE IT UNDER int main (){, constraint must be the same.

Answer 1

: : C code : :

#include <stdio.h>
int main()
{
  int lcm,sum,st,size,i,m,l,r,t,p;
  p=0;
  scanf("%d",&t);
  while(t>0)
  {
      p++;
      t--;sum=0;
      int arr[3];
      scanf("%d %d %d %d %d",&arr[0],&arr[1],&arr[2],&l,&r);
      size = 3;//sizeof(arr) / sizeof(int);
      lcm = arr[0];
      //calculating lcm of 3 numbers
      for (i = 1; i < size; i++){
        m=1;
        while( m%lcm || m%arr[i]) m++;
        lcm = m;
      }
      //finding starting index to start sum from for lcm multiple 
      if(l>= lcm) st=l/lcm;
      else st=1;
      
      for(i=st; lcm*i <= r ; i++ ){
          sum+=lcm*i;
      }
      printf("Case #%d: %d\n",p,sum);
  }return 0;
}

Algorithm

- We basically have to find the lcm of A,B,C ; first ,

-Then sum up multiples of lcm that lie in the range [L,R]

: : Sample output : :

Please refer to the screenshot of the code to understand the indentation of he code: