Hướng dẫn giải của Triomino Game
Chỉ dùng lời giải này khi không có ý tưởng, và đừng copy-paste code từ lời giải này. Hãy tôn trọng người ra đề và người viết lời giải.
Nộp một lời giải chính thức trước khi tự giải là một hành động có thể bị ban.
Nộp một lời giải chính thức trước khi tự giải là một hành động có thể bị ban.
Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này
Code mẫu của flashmt
#include <iostream> #include <algorithm> #include <set> #define fr(a,b,c) for (a=b;a<=c;a++) using namespace std; int g[4][888]; set <int> a; void init() { int i,j; fr(j,1,3) g[j][1]=1; fr(i,2,800) { // 0 a.clear(); fr(j,0,i-2) a.insert(g[0][j]^g[1][i-2-j]); for (j=0;a.count(j)>0;j++); g[0][i]=j; // 1 a.clear(); fr(j,0,i-1) a.insert(g[0][j]^(j+2<=i?g[2][i-2-j]:0)); fr(j,0,i-2) a.insert(g[0][j]^g[3][i-2-j]); fr(j,0,i-2) a.insert(g[1][j]^g[1][i-2-j]); for (j=0;a.count(j)>0;j++); g[1][i]=j; // 2 a.clear(); fr(j,0,i-1) a.insert(g[1][j]^(j+2<=i?g[2][i-2-j]:0)); fr(j,0,i-2) a.insert(g[1][j]^g[3][i-2-j]); for (j=0;a.count(j)>0;j++); g[2][i]=j; // 3 a.clear(); fr(j,0,i-1) a.insert(g[1][j]^(j+2<=i?g[2][i-2-j]:0)); fr(j,0,i-2) a.insert(g[1][j]^g[3][i-2-j]); for (j=0;a.count(j)>0;j++); g[3][i]=j; } } int main() { int n,test; init(); cin >> test; while (test--) { cin >> n; cout << (g[0][n]?'X':'Y') << endl; } return 0; }
Code mẫu của RR
{$R+,Q+} const FINP=''; FOUT=''; MAXN=801; oo=100; var f1,f2:text; test,n:longint; g:array[0..MAXN,0..1,0..1] of longint; dd:array[0..oo] of longint; procedure openF; begin assign(f1,FINP); reset(f1); assign(f2,FOUT); rewrite(f2); end; procedure closeF; begin close(f1); close(f2); end; var i:longint; procedure cal1(l:longint); begin fillchar(dd,sizeof(dd),0); for i:=0 to l-2 do dd[g[i,0,0] xor g[l-1-i,0,1]]:=1; for i:=0 to oo do if dd[i]=0 then begin g[l,0,0]:=i; exit; end; end; procedure cal2(l:longint); begin fillchar(dd,sizeof(dd),0); for i:=0 to l-2 do begin dd[g[i,0,0] xor g[l-1-i,1,1]]:=1; if i>0 then dd[g[i,0,1] xor g[l-1-i,0,1]]:=1; end; for i:=0 to oo do if dd[i]=0 then begin g[l,0,1]:=i; exit; end; end; procedure cal3(l:longint); begin fillchar(dd,sizeof(dd),0); for i:=1 to l-2 do begin dd[g[i,1,1] xor g[l-1-i,0,1]]:=1; dd[g[i,0,1] xor g[l-1-i,1,1]]:=1; end; for i:=0 to oo do if dd[i]=0 then begin g[l,1,1]:=i; exit; end; end; procedure init; var l:longint; begin g[2,0,0]:=1; g[2,0,1]:=1; for l:=3 to MAXN do begin cal1(l); cal2(l); cal3(l); end; end; begin openF; init; read(f1,test); for test:=1 to test do begin read(f1,n); if g[n,0,0]=0 then writeln(f2,'Y') else writeln(f2,'X'); end; closeF; end.
Code mẫu của hieult
#include <iostream> #include <cstdio> #include <set> //#include <conio.h> using namespace std; int dx[] = {0,0,1,1}; int dy[] = {0,1,0,1}; int triomino[1000][2][2],test,n; int main() { memset(triomino,0,sizeof(triomino)); for(int i = 1;i<=800;i++) { for(int j = 0;j<4;j++) { set <int> bo; set <int>::iterator it; if(dx[j]==0) bo.insert(triomino[i-1][1][dy[j]]); if(dy[j]==0) bo.insert(triomino[i-1][dx[j]][1]); for(int k = 0;k+2<=i;k++) { bo.insert(triomino[k][dx[j]][1]^triomino[i-k-2][0][dy[j]]); bo.insert(triomino[k][dx[j]][0]^triomino[i-k-2][1][dy[j]]); } for(int k = 0;;k++) if(bo.find(k)==bo.end()) { triomino[i][dx[j]][dy[j]]=k; break; } } } scanf("%d",&test); for(int i = 1;i<=test;i++) { scanf("%d",&n); if(triomino[n][1][1]) printf("X\n"); else printf("Y\n"); } // getch(); }
Code mẫu của ll931110
program TRIOMINO; const maxn = 800; maxv = 1000; var l: array[0..maxn,0..2,0..2] of integer; ex: array[0..maxv] of boolean; res: array[0..maxn] of boolean; procedure calc(x: integer); var i,q,t: integer; begin //Calc l[x,0,0] fillchar(ex, sizeof(ex), false); for i := 1 to x - 1 do begin q := l[i - 1,0,0] xor l[x - i,1,0]; ex[q] := true; end; t := 0; while ex[t] do inc(t); l[x,0,0] := t; //Calc l[x,0,1] fillchar(ex, sizeof(ex), false); for i := 1 to x - 1 do begin q := l[i - 1,0,0] xor l[x - i,2,1]; ex[q] := true; end; for i := 1 to x - 2 do begin q := l[i - 1,0,0] xor l[x - i,1,1]; ex[q] := true; end; for i := 2 to x - 1 do begin q := l[i - 1,0,2] xor l[x - i,0,1]; ex[q] := true; end; t := 0; while ex[t] do inc(t); l[x,0,1] := t; l[x,1,0] := t; l[x,0,2] := t; l[x,2,0] := t; //Calc l[x,1,1] fillchar(ex, sizeof(ex), false); for i := 2 to x - 2 do begin q := l[i - 1,1,0] xor l[x - i,1,1]; ex[q] := true; end; for i := 2 to x - 1 do begin q := l[i - 1,1,0] xor l[x - i,2,1]; ex[q] := true; end; t := 0; while ex[t] do inc(t); l[x,1,1] := t; l[x,2,2] := t; //Calc l[x,1,2] fillchar(ex, sizeof(ex), false); for i := 2 to x - 1 do begin q := l[i - 1,1,0] xor l[x - i,1,2]; ex[q] := true; end; for i := 2 to x - 2 do begin q := l[i - 1,1,0] xor l[x - i,2,2]; ex[q] := true; end; t := 0; while ex[t] do inc(t); l[x,1,2] := t; l[x,2,1] := t; end; procedure solve; var i,j: integer; begin fillchar(l, sizeof(l), 0); l[2,0,0] := 1; l[2,0,1] := 1; l[2,1,0] := 1; l[2,0,2] := 1; l[2,2,0] := 1; for i := 3 to maxn do calc(i); res[1] := false; res[2] := true; for i := 3 to maxn do if l[i,0,0] = 0 then res[i] := false else res[i] := true; end; procedure printresult; var i,k,n: integer; begin readln(k); for i := 1 to k do begin readln(n); if res[n] then writeln('X') else writeln('Y'); end; end; begin solve; printresult; end.
Code mẫu của skyvn97
#include<cstdio> const int res[]={1,0,0,1,0,0,1,0,0,0,0,1,0,0,0,1,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}; int n,c,t; int main(void) { scanf("%d",&t); for (c=1;c<=t;c=c+1) { scanf("%d",&n); printf("%c\n",res[n-1]+'X'); } return 0; }
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