HDU5090 simulation, hash

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/* HDU 5090 is a simple simulation question, however, there is a deep hash idea. this is my first hash question. it is also my first paper code, without compilation or running. after so many comments, it is worth remembering., because n is very small and does not exceed 100, you can open an array to record how many times each number appears because I + n * k can only increase, therefore, you only need to check from 1 to n one by one. if the hash [I] of the current check is 0, there is no solution: because it is impossible to change other numbers to the hash [I]> 1 of the current check, the I must be changed to j = n * k + I (n> 0 ), hash [j] = 0 indicates that j has not changed from number I in the input number. if the hash [I] = 1 of the current check is repeated, you only need to determine the flag */# include
 
  
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           Using namespace std; # define input freopen ("input.txt", "r", stdin); # define output freopen ("output.txt", "w", stdout ); # define For1 (I, a, B) for (I = a; I
          
            B; I --) # define Dec2 (I, a, B) for (I = a; I> = B; I --) # define Sca_d (x) scanf ("% d", & x) # define Sca_s (x) scanf ("% s", x) # define Sca_c (x) scanf ("% c ", & x) # define Sca_f (x) scanf ("% f", & x) # define Sca_lf (x) scanf ("% lf", & x) # define Fill (x, a) memset (x, a, sizeof (x) # define MAXN 1110 template
           
             T gcd (T a, T B) {return B = 0? A: gcd (B, a % B);} template
            
              T lcm (T a, T B) {return a/gcd (a, B) * B;} int main () {int hash [MAXN]; int t, n, k, I, j, m, flag; cin> t; while (t --) {cin> n> k; Fill (hash, 0); For2 (I, 1, n) Sca_d (m), hash [m] ++; flag = 1; For2 (I, 1, n) if (! Hash [I]) {flag = 0; break;} else if (hash [I] = 1) continue; else {for (j = I + k; j <= n; j + = k) if (hash [I]> 1 & hash [j] = 0) {hash [j] = 1; hash [I] -- ;}}if (flag) cout <"Jerry \ n"; else cout <"Tom \ n" ;}return 0 ;}