题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4952

Number Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 261    Accepted Submission(s): 117

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

官方题解：http://blog.sina.com.cn/u/1809706204

打表找规律也行！

当时在打表的时候10W是WA,20W是T，最后12W才过！

代码如下：

#include<cstdio>
typedef __int64 LL;
int main()
{
    int cas=1;
    LL i;
    LL x,k;
    while(scanf("%I64d %I64d",&x,&k)!=EOF)
    {
        if(x==0 && k==0)
            break;
        if(k <= 120000)
        {
            for(i = 1; i <= k; i++)
            {
                if(x%i==0)
                    continue;
                else
                    x=(x/i+1)*i;
            }
            printf("Case #%d: ",cas++);
            printf("%I64d\n",x);
        }
        else
        {
            LL y1,y2;
            for(i = 1; i <= 120000; i++)
            {
                if(x%i==0)
                    continue;
                else
                    x=(x/i+1)*i;
            }
            y1=x;
            y2=(x/i+1)*i;
            printf("Case #%d: ",cas++);
            printf("%I64d\n",y1+(k-120000)*(y2-y1));
        }
    }
    return 0;
}

hdu4952 Number Transformation（数学题 | 找规律）