Number Transformation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 171 Accepted Submission(s): 69

Problem Description

Teacher Mai has an integer x.

He does the following operations k times. In the i-th operation, x becomes the least integer no less than x, which is the multiple of i.

He wants to know what is the number x now.

Input

There are multiple test cases, terminated by a line "0 0".

For each test case, the only one line contains two integers x,k(1<=x<=10^10, 1<=k<=10^10).

Output

For each test case, output one line "Case #k: x", where k is the case number counting from 1.

Sample Input

2520 10

2520 20

0 0

Sample Output

Case #1: 2520

Case #2: 2600

Source

2014 Multi-University Training Contest 8

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    __int64 x,i,k,j;
    int t=1;
    while(~scanf("%I64d%I64d",&x,&k))
    {
         if(x==0&&k==0)
            break;
         i=2;
         while(i<=k)
         {
             if(x%i!=0)
             {
                j=x/i+1;//求倍数。
                  if(j<=i)
                    {
                        x=k*j;
                        break;
                    }
                x=j*i;
             }
             else
              {
                 j=x/i;
                 if(j<=i)
                    {
                        x=k*j;
                        break;
                    }
              }
             i++;
         }
      printf("Case #%d: %I64d
",t++,x);
    }
    return 0;
}