Number Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 514    Accepted Submission(s): 260

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 102520 200 0

 

Sample Output

Case #1: 2520Case #2: 2600

 

Source

2014 Multi-University Training Contest 8

这题目出的，让我都不知道怎么用汉语表达了

题意：输入两个数，x,k.要保证从1~k一个一个都能被整除。并且要保证这个数据比前面x要大或者相等

例如:2520 20 前面10位都能被2520整除，当i变到11.x就会变成2530.2530/11=230；并且2530》2520；

分析：设第i个数为i*a;第i+1个数为（i+1）*b。则（i+1）*b>i*a;b>a-(a/(i+1));那么如果a<i+1;b的值就会不变

此时k*b就为所求；而由于a在不断的变小，a<i+1;在数值比较大的情况下就能实现；

下面看代码实现：

#include<stdio.h>int main(){    int t=0;    __int64 a,b,i,j;    while(scanf("%I64d%I64d",&a,&b),a||b)    {        t++;      for(i=2;i<=b;i++)      {          if(a%i!=0)          {              j=a/i+1;             a=i*j;//普通情况下的求解                }          else                {                    j=a/i;//计算出j的数值，便于判断。                }                if(j<i)                {                    a=j*b;                    break;//当数值满足j<i;就会跳出：                }    }    printf("Case #%d: %I64d\n",t,a);    }    return 0;}