bzoj 3858: Number Transformation 暴力

3858: Number Transformation

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 82  Solved: 41
[Submit][Status]

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

HINT

Source

　　其实我也不知道怎么证明，反正通过实验发现n/i每次严格减小到一个极限值，就再也不会变了，而这个极限值在sqrt(n)范围，迭代次数也是sqrt(n)的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long qword;
int main()
{
        freopen("input.txt","r",stdin);
        qword n,m;
        int i;
        qword t;
        int cnt=0;
        while (scanf("%lld%lld",&n,&m),cnt++,n+m)
        {
                t=0;
                for (i=1;i<=m;i++)
                {
                        n=(n/i+(n%i!=0))*i;
                        if (n/i==t)break;
                        t=n/i;
                }
                printf("Case #%d: %lld
",cnt,m*t);
        }
}