UVA 11549 - Calculator Conundrum Floyd判圈法

Problem C

CALCULATOR CONUNDRUM

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

 
  OUTPUT 
  
9
99

Calgary Collegiate Programming Contest 2008


Floyd判圈法



#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#endif


const int INF = 0x3f3f3f3f;
const int maxn = 100 + 10;

int buf[20], bi;
int next(int n, int k) {
	LL t = (LL) k * k;
	bi = 0;
	while(t) {
		buf[bi++] = t % 10;
		t /= 10;
	}
	int res = 0;
	for(int i=bi-1; i>=max(0, bi-n); i--) {
		res = res * 10 + buf[i];
	}
	return res;
}

int main() {
	int T;

	scanf("%d", &T);
	while(T--) {
		int n, k;
		scanf("%d%d", &n, &k);
		int ans = k;
		int k1=k, k2=k;
		do {
			k1 = next(n, k1);
			k2 = next(n, k2); if(k2 > ans) ans = k2;
			k2 = next(n, k2); if(k2 > ans) ans = k2;
		} while(k1 != k2);
		printf("%d\n", ans);
	}

	return 0;
}




你可能感兴趣的:(ACM)