UVA 11549 - Calculator Conundrum Floyd判圈法

Problem C

CALCULATOR CONUNDRUM

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

She enters a numberk then repeatedly squares it until the result overflows. When the result overflows, only then 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:

“Givenn 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 integersn (1 ≤ n ≤ 9) and k (0 ≤ k < 10ⁿ) wheren 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

21 62 99

OUTPUT

999

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Calgary Collegiate Programming Contest 2008

Floyd判圈法

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <string>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <set>using namespace std;#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#endifconst 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;}