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HDU-4952 Number Transformation
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): 171 Accepted Submission(s): 69
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<iostream>#include<cstdio>using namespace std;int main(){ __int64 x,i,k,j; int t=1; while(~scanf("%I64d%I64d",&x,&k)) { if(x==0&&k==0) break; i=2; while(i<=k) { if(x%i!=0) { j=x/i+1;//求倍数。 if(j<=i) { x=k*j; break; } x=j*i; } else { j=x/i; if(j<=i) { x=k*j; break; } } i++; } printf("Case #%d: %I64d\n",t++,x); } return 0;}
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