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hdu4952Number Transformation (找规律)
Number Transformation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 76 Accepted Submission(s): 28
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.
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).
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<stdio.h>
int main()
{
__int64 x,k,i,j,t=0;
while(scanf("%I64d%I64d",&x,&k)>0&&(x||k))
{
i=2;
while(i<=k)
{
if(x%i!=0)
{
j=x/i+1; if(j<=i){x=j*k;break;}
x=i*j;
}
else
{
j=x/i; if(j<=i){x=j*k;break;}
}
i++;
}
printf("Case #%I64d: %I64d\n",++t,x);
}
}
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