Zoj 1039 number game (State compression, memory-based search)

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Question: 2-20 this 19-digit game. Each time a number is taken away, the number multiples cannot be obtained, and the sum of two numbers that cannot be obtained.

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 39

Because there are only 19 data records, you can perform a memory search by compressing and storing the status. DP [1 <19]. DP [I] indicates that a viable set of I is a must-win situation.

Next, state transfer is important.

For the original state, if you want to remove the number X from the set. There are two steps:

1. Remove the multiples of x and X. This is to perform a bit operation at a specific position of a number 0. For more information, seeCode

2. Remove the sum of elements not in the set and multiples of X. This step can be implemented in reverse order to find the elements in the set and subtract several X numbers from them to determine whether the new number is in the set. If not, this number is invalid, remove. It can also be solved through bitwise operations

In winning and losing, the N-State must follow the p-state.

Finally, we need to output a feasible solution, which is to enumerate and delete a certain number and determine whether to change to the P state.

# Include <iostream> # include <cstdio> # include <ctime> # include <cstring> # include <cmath> # include <algorithm> # include <cstdlib> # include <vector> # Define c 240 # define time 10 # define INF 1 <25 # define ll long longusing namespace STD; int DP [1 <19]; int get_state (INT state, int X) {int ret = State; // remove the multiples of x and X. // first, set the location to 1, and then reverse the value. Finally, it matches the state phase with for (INT I = X; I <= 20; I + = x) RET & = ~ (1 <(I-2); For (INT I = 2; I <= 20; I ++) {// If a number is still in the set, then determine whether there is a number not in the Set and the multiples of X constitute if (1 <(I-2) & RET) {for (Int J = x; i-j-2> = 0; J + = x) // If I is the sum of a number not in the set and X, if (! (1 <(i-j-2) & RET) {RET & = ~ (1 <(I-2); break ;}} return ret ;}int get_dp (INT state) {If (DP [State]! =-1) return DP [State]; for (INT I = 2; I <= 20; I ++) {If (State & (1 <(I-2 ))) {// remove the number of I from the set int TMP = get_state (state, I); // P State to obtain the N state if (! Get_dp (TMP) return DP [State] = 1 ;}} return DP [State] = 0 ;}int main () {memset (DP,-1, sizeof (DP); DP [0] = 0; int t, n, a [20], CAS = 0; scanf ("% d", & T ); while (t --) {scanf ("% d", & N); int state = 0; For (INT I = 0; I <n; I ++) {scanf ("% d", & A [I]); // initial state | = 1 <(A [I]-2 );} printf ("Scenario # % d: \ n", ++ CAS); If (get_dp (state) = 0) puts ("There is no winning move. \ n "); else {printf (" the winning moves are: "); For (INT I = 0; I <n; I + +) {// Enumerative removal, whether it is a mandatory defeat state int TMP = get_state (State, a [I]); If (! Get_dp (TMP) printf ("% d", a [I]);} puts (". \ n") ;}} return 0 ;}