本文主要是介绍HDOJ 4952 Number Transformation，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目：
Number Transformation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 209 Accepted Submission(s): 91
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
解题思路：
通过打表，我们可以发现这题其实是有规律的，假设第i次时满足题意的值为ai，当ai/i=a(i+1)/(i+1)时，以后所有的数都满足a(t)/t = a(i)/i,我们可以直接算出最后的结果ai*k/i。
代码：
#include <cstdio>
#include <cstring>typedef long long lint;
lint x, k, ans;int main()
{int icase = 1;while(~scanf("%I64d%I64d", &x, &k)){if(0==x && 0==k) break;lint b = 0; ans = x;// lint w = 0;bool flag = false;for(lint i=1; i<=k; ++i){if(0 == ans%i) continue;lint c = ans / i;// printf("%I64d %I64d %I64d\n", i, b, c);if(b == c+1){flag = true;// w = i;break;}ans = (c+1) * i;b = c + 1;}if(flag) ans = k * b;// printf("%I64d %I64d ", b, w);printf("Case #%d: %I64d\n", icase++, ans);}return 0;
}
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