Calgary Collegiate Programming Contest 2008 / UVa 11549 Calculator Conundrum (Floyd判圈算法&更快的优化)

11549 - Calculator Conundrum

Time limit: 6.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=460&page=show_problem&problem=2544

Alice got a hold of an old calculator that can display n digits. She was bored enough to come up with the following time waster.

She enters a number k then repeatedly squares it until the result overflows. When the result overflows, only the n most significant digits are displayed on the screen and an error flag appears. Alice can clear the error and continue squaring the displayed number. She got bored by this soon enough, but wondered:

“Given n and k, what is the largest number I can get by wasting time in this manner?”

Program Input

The first line of the input contains an integer t (1 ≤ t ≤ 200), the number of test cases. Each test case contains two integers n (1 ≤ n ≤ 9) and k (0 ≤ k < 10ⁿ) where n is the number of digits this calculator can display k is the starting number.

Program Output

For each test case, print the maximum number that Alice can get by repeatedly squaring the starting number as described.

Sample Input & Output

INPUT

2
1 6
2 99

OUTPUT

9
99

首先按题意这么算，肯定会出现循环。

如何判断刚好出现循环的位置？——Floyd判圈算法。

完整代码：

/*0.212s*/

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;

LL n, k, base;

LL count(LL x)///取前n个数的优化写法
{
	while (x >= base)
		x /= 10;
	return x;
}

int main(void)
{
	int cas;
	scanf("%d", &cas);
	while (cas--)
	{
		LL ans, k1, k2;
		scanf("%lld%lld", &n, &k);
		base = 1;
		while (n--)
			base *= 10;
		k1 = k2 = k;
		ans = k;
		do
		{
			k1 = count(k1 * k1);
			k2 = count(k2 * k2);ans = max(ans, k2);
			k2 = count(k2 * k2);ans = max(ans, k2);
		}
		while (k1 != k2);
		printf("%lld\n", ans);
	}
}