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 < 10 n ) where n is the number of digits this calculator
can display k is the starting number.
Output
For each test case, print the maximum number that Alice can get by repeatedly squaring the starting
number as described.
Sample Input
2
1 6
2 99
Sample Output
9
99
数字肯定最后肯定会重复，floyd判圈算法，节省了set的空间开销
#include <cstdio> #include <algorithm> #include <iostream>using namespace std;int n;int next(long long x) {int y = 0;x *= x;for (long long t = x; t; t /= 10, ++y);if (n < y)for (int i = 0; i < y - n; ++i) x /= 10;return x; }int main() {int T, k;cin >> T;while (T--) {cin >> n >> k;//注意题给的条件，n,k的关系int t1, t2, m;t1 = t2 = m = k;do {t1 = next(t1);t2 = next(t2);m = max(m, t2);t2 = next(t2);m = max(m, t2);} while (t1 != t2);cout << m << endl;} }
)
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