Calgary Collegate Programming Contest 2008/UVa 11549 Calculator Conundrum(Floydコイルアルゴリズム&より高速な最適化)

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

まず題意通りに計算すると、必ずループが現れます.
ちょうどループが発生した位置をどう判断しますか.--Floydコイルアルゴリズム.
完全なコード:

/*0.212s*/

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

LL n, k, base;

LL count(LL x)///  n       
{
	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
", ans);
	}
}