Number Transformation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 617 Accepted Submission(s): 313
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
题解及代码：
这道题目在比赛时是打表看出来的规律，我把i，n/i，n暴力输出出来，看到当i*i>=n时，n/i不再发生变化，由于数据最大是10^10，所以我们每次最多计算10^5次，不会超时，因此我们每次最多计算到i*i>=n时就可以了。
#include
#include
using namespace std; int main() { long long n,k; int cas=1; while(scanf("%I64d%I64d",&n,&k)!=EOF) { if(!n&&!k) break; for(long long i=1;i<=k;i++) { if(n%i) { long long t=n/i; t++; n=t*i; } if(i*i>=n) { n=(n/i)*k; break; } } printf("Case #%d: %I64d\n",cas++,n); } return 0; }