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给出四个数字,要求,在其间添加运算符和括号,使得计算结果等于24。5 h6 w p: e" M: s1 U8 {1 C
* V! o4 U* V* `& D+ } 括号的放置即为决定哪几个数先进行计算。所以,我们先确定首先进行计算的两个相邻的数,计算完成后,就相当于剩下三个数字,仍需要在它们之间添加符号;然后再决定在这三个数中哪两个相邻的数先计算。由此,我们就成功解决了数字的运算次序问题,此时不需要再考虑不同运算符号的优先级问题,因为括号的优先级高于加减乘除。
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通过循环,我们可以得到第一第二第三次计算的运算符,再通过计算,就可以得出和,若和等于24,即为所求解。# h& J# n9 e2 B3 V: G& h, R1 @6 ^+ W
9 ]5 S9 i5 Q) [/ K4 d# @: W 在输出格式中,由于括号的放置共六种情况,故根据计算先后顺序的不同,输出时在不同地方放置括号;- r5 o/ R1 M! e' I' [
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以下是java 源码 :# t+ R. A9 n) n9 j
import java.awt.*;
- ^% [. X x" Y& X, f import javax.swing.*;4 L k, @1 _& N: }
import java.awt.event.*;
! u, J% D" v# e* Z |( Y import java.util.*;
6 l0 Z$ t9 t. V import javax.swing.JOptionPane;! ^/ k& h! e' C' G h2 |: p5 |
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public class TwentyFourPoke_Game extends JFrame. G: q2 H- g- f) `. m5 |0 ?
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private JButton jbsolution = new JButton("Find a Solution");
1 D9 Q# |( w+ r/ d1 y( y private JButton jbrefresh = new JButton("Refresh");1 N1 [* [; Y# n2 a
private JButton jbverify = new JButton("Verify");" P8 i/ J. X3 S$ x
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private JLabel jlmessage = new JLabel("Enter an expression:");
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private JTextField jtsolution = new JTextField();
c7 Y z7 z# X7 @& U6 v6 A4 a private JTextField jtexpression = new JTextField();
4 f- {& ^8 r7 ?
( v/ a! y$ p( a; L C" N6 B; p private ImagePanel pp = new ImagePanel();- R3 X7 w W; z; s9 E) J& a4 |
+ x S! I5 A) Y6 C8 X$ n) W private int[] card = new int[4];
- K5 b, N/ b3 Y private double[] bcard = new double[4];+ j" {3 e1 T) }/ h b4 u4 X
" H# v8 Z" V H8 P+ ~6 A private double sum;* I/ I g* a/ V2 [# ?; S/ r" b
private double[] temp1 = new double[3];- q! @0 t! \" v$ B
private double[] temp2 = new double[2];
/ Y" P3 B) w1 |& n4 x, R# q% D5 L6 I: A- u private char[] sign = {'+','-','*','/'};
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& y0 J, k7 f. d5 G) f: D public TwentyFourPoke_Game(), r) v3 Z$ N* ]7 Z0 f
{
+ [/ ^ V, z/ I3 U {" i JPanel p1 = new JPanel(new GridLayout(1,3));* e7 y; M, R: P1 B3 V% f* I8 E& |
p1.add(jbsolution);
+ J4 N! b2 v3 e$ v5 h9 { p1.add(jtsolution);
" |; M! |' H0 B( w8 g! A p1.add(jbrefresh);" R3 c8 k. M% q
JPanel p3 = new JPanel(new GridLayout(1,3));
( {/ ]0 @/ s0 \/ P+ H p3.add(jlmessage);% a; @& @! M& j9 P" [8 m/ @$ a
p3.add(jtexpression); V; n2 R5 m6 Q& a4 L) i) N5 o
p3.add(jbverify); y7 G/ q0 w% J
4 m9 }. }# D4 R4 |) S3 w& P, {
add(p1,BorderLayout.NORTH);
$ a! w: o: w# P2 p( [7 U' U9 r add(pp,BorderLayout.CENTER);
/ u0 Q0 w+ b% O& P2 S. w5 i* y. C add(p3,BorderLayout.SOUTH);, c; |. _3 Y! a' b/ n5 g: @: D$ R
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ButtonListener listener = new ButtonListener();: _1 B3 H; Y B$ k6 R
jbsolution.addActionListener(listener);
4 B0 y0 C: X* Z7 V5 c5 z/ K2 d jbrefresh.addActionListener(listener);
& W( N5 P/ C" Q# D# E' n8 x jbverify.addActionListener(listener);
$ O2 d1 v7 f0 n- F8 U }9 \" _7 b( l# h: s9 o9 ^, _
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class ButtonListener implements ActionListener& d r/ }# l; H; {
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public void actionPerformed(ActionEvent e)8 n* J1 F6 Y/ I7 |+ ?
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if(e.getSource() == jbsolution)
: g F! f3 ~" |/ j! M" u) D$ Z {
$ _ w2 |( R' U) s for(int i = 0;i < 4;i++)( F5 e% F: t9 w) N o, Z' a
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bcard[i] = (double)card[i] % 13;
3 Q, Q# N6 l6 b! { if(card[i] % 13 == 0)3 A; L& K* L, P# V% l
bcard[i] = 13;
- P2 d# E' a+ Y1 z$ z/ P }
S& s( C) t3 c7 s9 \- r search();/ \: g! _2 @. E5 C0 k
}
3 [ O- f* d1 ` else if(e.getSource() == jbrefresh)) }% w2 M& y, a/ O
{9 C) o1 X7 r% [8 m6 g! \
pp.sshow();
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else if(e.getSource() == jbverify)+ k2 J' T. r/ h6 p2 O1 r
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String expression = jtexpression.getText();
# j" N7 x1 ?$ X" [2 r5 y, {- O int result = evaluateExpression(expression);
' p4 U( B( Y) A* p0 B* N if(result == 24)
9 M, y- D. P- x8 F {
$ L; B$ y5 ~# q7 W JOptionPane.showMessageDialog(null,"恭喜你!你答对了!","消息框",JOptionPane.INFORMATION_MESSAGE);3 d( L6 S0 |4 W- I9 ~
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else. K9 z' r2 j/ X' ^( q5 d
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JOptionPane.showMessageDialog(null,"抱歉!请再次尝试。","消息框",JOptionPane.INFORMATION_MESSAGE);0 A9 [/ V1 n6 @4 k& r2 A
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}
+ u( ^- V; _" {' O }
" T+ e8 l! d$ `- T5 p A# C
* H" Q# Z9 g- M; W0 s. f public static double calcute(double a,double b,char c)9 I: P, z; M4 A% t
{
% f) z# F% p3 }# K- ^& C if(c == '+'), G6 ^6 P$ b% A- @
return a+b;
+ k/ i4 e/ b: \* D3 W else if(c == '-')
8 Q% H- I( p$ a" Z& g0 h return a-b;
+ j0 F' F' q! U! f else if(c == '*') Y- i7 Q9 E1 @
return a*b;
$ g9 T, {1 t8 Z p6 b5 e else if(c == '/' && b != 0) n, B0 I c9 M2 P \2 {' y5 j
return a/b;
' {4 `: Q b- b1 u+ h- u& e else
9 S q9 j2 Z+ \4 c H* d* u/ F return -1;
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/ z/ F9 t, B3 p% S' f. X$ `! L) e4 Q public void search()5 o) L2 X9 l) t. z( N. e' f
{
/ `( R! p; d4 E- F9 _8 l6 w boolean judge = false;( t7 W: \6 o+ X' H2 m- f
for(int i=0;i<4;i++)- O ]) v) p8 R; R
//第一次放置的符号
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+ ~: Q6 G$ n# h( j" l3 P6 @ for(int j=0;j<4;j++)* _6 T5 {: ]1 A
//第二次放置的符号
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for(int k=0;k<4;k++)
4 w+ D e' f5 [7 L8 R& S //第三次放置的符号
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for(int m=0;m<3;m++)
+ s" V _+ g% | //首先计算的两个相邻数字,共有3种情况,相当于括号的作用! E m+ c( }( _
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if(bcard[m+1]==0 && sign[i]=='/') break;
% H. I% \; G9 L4 A. i temp1[m]=calcute(bcard[m],bcard[m+1],sign[i]);' `/ M' I8 a2 t" L
temp1[(m+1)%3]=bcard[(m+2)%4];+ {; U. }, s+ q( {# F, K& ?
temp1[(m+2)%3]=bcard[(m+3)%4];. B3 S$ A( T1 @. G* [
//先确定首先计算的两个数字,计算完成相当于剩下三个数,按顺序储存在temp数组中
% V% z0 R9 @( D2 d. i) Z for(int n=0;n<2;n++)$ m! D3 a* u9 v3 V
//三个数字选出先计算的两个相邻数字,两种情况,相当于第二个括号- i/ Z/ a# v4 \
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if(temp1[n+1]==0 && sign[j]=='/') break;
" g6 {1 g5 m' e5 v, ^" A temp2[n]=calcute(temp1[n],temp1[n+1],sign[j]);
) P$ ?& Q6 A6 ^% @# C4 i m temp2[(n+1)%2]=temp1[(n+2)%3];
8 L4 p& f: ?+ p( r9 r. k$ U# | //先确定首先计算的两个数字,计算完成相当于剩下两个数,按顺序储存在temp数组中
6 u2 E; Q; a: P/ H: E9 e if(temp2[1]==0 && sign[k]=='/') break;3 f* `5 N: b7 M5 U: n
sum=calcute(temp2[0],temp2[1],sign[k]);4 s9 v c+ `7 e$ U9 z9 l c; e
//计算和
8 M4 q, w% z! k' f- _ if(sum==24)8 `9 ?: i- Y/ N( H
//若和为24
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4 C4 y/ D J6 n6 B2 `! m( R judge=true;7 U* t$ z) W# @8 ?" `& Q- r! W
//判断符为1,表示已求得解, w9 p/ ]# L$ b
if(m==0 && n==0)
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String sss ="(("+(int)bcard[0]+sign[i]+(int)bcard[1]+")"+sign[j]+(int)bcard[2]+")"+sign[k]+(int)bcard[3]+"="+(int)sum;0 O& Q p# k6 E1 M: j
jtsolution.setText(sss);% B* k- W' p3 S2 e
return ;6 h0 Q3 z. m9 L
} e N# e6 x" S4 g, t
else if(m==0 && n==1)
; r- B1 e. u9 C {
! @- j j7 s+ m5 j+ C3 u; U String sss ="("+(int)bcard[0]+sign[i]+(int)bcard[1]+")"+sign[k]+"("+(int)bcard[2]+sign[j]+(int)bcard[3]+")="+(int)sum;) S7 @7 ?4 Q- m, E
jtsolution.setText(sss);
. ~3 T& S9 i! V m* A) G return ;
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else if(m==1 && n==0)" _/ k8 i0 [/ q: R3 C9 E
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String sss ="("+(int)bcard[0]+sign[j]+"("+(int)bcard[1]+sign[i]+(int)bcard[2]+"))"+sign[k]+(int)bcard[3]+"="+(int)sum;" U1 E3 K! } @: e, C) c
jtsolution.setText(sss);
; y* X# d% A: Z! e U9 t7 g7 ? return ;
- o) x) Y" B* g1 Y3 H& R, x }
! F/ d0 C" |, Q7 Z/ T% |* t else if(m==2 && n==0)
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String sss ="("+(int)bcard[0]+sign[j]+(int)bcard[1]+")"+sign[k]+"("+(int)bcard[2]+sign[i]+(int)bcard[3]+")="+(int)sum;9 g3 j; W8 U; I$ F
jtsolution.setText(sss);, }/ _6 \3 O& o/ ~# L/ T
return ;. r+ K& }1 [) ` U B9 Y& Z
}# S4 i7 C, T) ]3 g" C `! r
else if(m==2 && n==0)
4 ]; b; p! M6 w4 l. Q: o$ d {
7 I" q2 B% R9 E9 { String sss =(int)bcard[0]+sign[k]+"("+(int)bcard[1]+sign[j]+"("+(int)bcard[2]+sign[i]+(int)bcard[3]+"))="+(int)sum;
* V& s/ E: W. R- N: W+ \' o jtsolution.setText(sss);
% h' I9 h# o6 c$ a return ;+ J% y: a+ {/ v* {# C0 C; A9 k- \
}
% ]. R# \$ Y2 {* Q5 t //m=0,1,2 n=0,1表示六种括号放置可能,并按照这六种可能输出相应的格式的计算式7 F/ k' A: F0 N4 V7 w
4 D8 P( `& N: k: ?. x }" g, a& l5 @' I& O2 {% a3 ^
}3 d$ W0 z/ u/ j% o8 A R' ]4 b8 n
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}
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if(judge==false)! C- L" v3 H d
jtsolution.setText("No solution!");
: g( ~. }. m$ ^ //如果没有找到结果,符号位为0 x8 ?* `2 x2 ?) b' G/ I
}. }* I' B" V5 r; ~& t+ x: g
+ L! i0 k6 l- K* b/ [ 9 j5 s, Z" K! [1 G1 W$ {
public static int evaluateExpression(String expression), Y# j& h0 ^1 ]1 k& c, ]
{
0 m6 `# n! f# D( P0 X1 x& r0 h // Create operandStack to store operands
$ g( \+ D% O* K' M4 k java.util.Stack operandStack = new java.util.Stack();8 T3 z( a# O, D: P
7 P. n( Z9 O1 R) S( F( y // Create operatorStack to store operators2 t6 B2 t; l8 N& g. O
java.util.Stack operatorStack = new java.util.Stack();) N% ?: {0 P+ g
1 G! \4 B8 z5 r# T/ z // Extract operands and operators
' e' d! j4 h1 _8 M" Z; ` java.util.StringTokenizer tokens = new java.util.StringTokenizer(expression, "()+-/*", true);
$ i/ f& \: \; [# Z9 o8 _# R
. w# C' U" |/ q# {% G2 \' y // Phase 1: Scan tokens$ }! c& o, j& x6 v" Y* F) Z( ^
while (tokens.hasMoreTokens())4 Y( L, i4 x# R3 M
{
3 {: K8 Q) z/ P String token = tokens.nextToken().trim(); // Extract a token
# A) n- e" G5 y: g% u if (token.length() == 0) // Blank space
$ a: }+ g% j! ?4 R continue; // Back to the while loop to extract the next token; G w0 p& t$ q; x
else if (token.charAt(0) == '+' || token.charAt(0) == '-'), L: H6 G4 o* ^ H
{
e! R0 G+ a; k- ^ // Process all +, -, *, / in the top of the operator stack
) a; V+ |2 s, V$ o" u" P+ d6 z( A# c3 G while (!operatorStack.isEmpty() &&(operatorStack.peek().equals('+') ||operatorStack.peek().equals('-') || operatorStack.peek().equals('*') ||1 ^0 j; s: | V N
operatorStack.peek().equals('/'))) v5 c! J) b$ U8 y
{! A3 x5 s% w4 h( V7 F( G
processAnOperator(operandStack, operatorStack);
7 `- L2 x6 b6 ]+ w& b: ^ }
/ }+ ]( n( u# P# W$ Q4 W3 D // Push the + or - operator into the operator stack
. x5 \ g3 g0 s4 E operatorStack.push(new Character(token.charAt(0)));
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else if (token.charAt(0) == '*' || token.charAt(0) == '/')9 a8 ? c1 ]+ O T7 t) C
{
- D( B' _1 d2 U g1 B I' X // Process all *, / in the top of the operator stack
" U ^) U# c8 u( p while (!operatorStack.isEmpty() && (operatorStack.peek().equals('*') || operatorStack.peek().equals('/')))
6 V7 P! l# f% ~( Z* V {
* Z5 v% [) }' d$ E processAnOperator(operandStack, operatorStack);
- u! N- r7 l+ m }
* L6 x: J* ]& S* I " d' u# u: p1 T: B) P& e
// Push the * or / operator into the operator stack
0 }! i/ \! ]. L operatorStack.push(new Character(token.charAt(0)));
5 Y9 A% g: k" \6 I- k1 N) f" q }
" s- I5 f( g! r$ B else if (token.trim().charAt(0) == '(')8 }6 w( M) w4 C* p7 I! k
{
) C# l& o! T9 K0 X4 i operatorStack.push(new Character('(')); // Push '(' to stack5 r4 ^; I6 k0 t' Y# c( C1 l
}' E& `0 r7 ~; f- i4 T
else if (token.trim().charAt(0) == ')')6 M; L+ [ i9 l4 [( \. l' N
{+ D9 M' q2 S; G0 P7 n7 m/ Y/ r
// Process all the operators in the stack until seeing '('- G0 s2 |' W- W0 S/ @8 w: L" F
while (!operatorStack.peek().equals('('))
2 q0 x5 P n0 M g! z2 S3 @ {
5 _6 z2 [+ g5 g3 Y processAnOperator(operandStack, operatorStack);- B3 e/ r" T2 _; W% b2 P* L
}. Y3 q8 n$ y/ [$ X
operatorStack.pop(); // Pop the '(' symbol from the stack% f6 m5 p1 U( a# _7 |6 I
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else
4 h& w7 G4 H2 o* U$ {" w! V {$ f( S4 |, ~; `- I
// An operand scanned9 g) t# Y% k3 Q/ ^
// Push an operand to the stack
0 r \. g1 t9 g! x) j! _% k operandStack.push(new Integer(token));
' ~" H4 g: n- t }7 F- Z) o& c7 I( F4 W! O4 u
}
: D) X) U0 @4 `& S' t) @ ) H% S3 r- _4 ?
// Phase 2: process all the remaining operators in the stack
* r% j- Q+ c7 s1 Y- Y9 B while (!operatorStack.isEmpty())! s6 w1 B& Q. T' I _
{
2 W; D( p# y6 [4 Y" _! k6 ^ processAnOperator(operandStack, operatorStack);4 \+ q+ a- t4 @- A
}7 E$ K( g) `+ P7 @$ t4 g: X
+ _5 Z+ u' k7 d8 t( Y) d // Return the result
; c/ s% X- i' W# T- T: Z* G return ((Integer)(operandStack.pop())).intValue();8 C/ @- d+ |1 u8 i1 a
}
6 ^# f. K1 V* u3 [3 l* f
& B' E5 H- @% Q public static void processAnOperator(java.util.Stack operandStack,java.util.Stack operatorStack)
9 I% y$ `, H. m! J* H {5 L4 D. |. i/ [
if (operatorStack.peek().equals('+'))
& h' `) k: F% e" u; j- L {
s! X8 y7 T* \$ c/ E& F operatorStack.pop();8 w" ]+ z/ M" d
int op1 = ((Integer)(operandStack.pop())).intValue();, [+ p6 L, r" D3 T! X
int op2 = ((Integer)(operandStack.pop())).intValue();
* Z. K) z" I5 j& C0 D0 X operandStack.push(new Integer(op2 + op1));% p8 y- d6 B$ I1 E, w+ L
}/ O% H# Z |0 b
else if (operatorStack.peek().equals('-'))
9 D; K ?' Z$ T: e3 P9 V" P! G- I {
* s/ N* }7 E) m7 t operatorStack.pop();
2 |) t4 ]; p% H int op1 = ((Integer)(operandStack.pop())).intValue();
8 \1 V/ s% G" U) W int op2 = ((Integer)(operandStack.pop())).intValue();
1 W6 m8 D: B0 T8 y3 v9 u; Y; b! ? operandStack.push(new Integer(op2 - op1));
! \* Y O. F. z: l& h/ I0 ` }7 P0 r \9 T& g# N0 g
else if (operatorStack.peek().equals('*'))
7 V$ J* a4 j7 E( H {$ u/ ]. V3 g8 T
operatorStack.pop();
' N, w" z! ]2 @9 | int op1 = ((Integer)(operandStack.pop())).intValue();1 b* j. v, D# S& _
int op2 = ((Integer)(operandStack.pop())).intValue();
+ e3 e$ h6 Z( r; S; x operandStack.push(new Integer(op2 * op1));
$ D8 |2 W* o" a; V8 W9 Q } ]+ Z- F3 V, l$ N+ c* _
else if (operatorStack.peek().equals('/'))' S" ]! j$ T) D7 H
{
6 D" C0 y" ~) a6 m$ R% { operatorStack.pop();; v# b% q: X2 G% R' L; y
int op1 = ((Integer)(operandStack.pop())).intValue();8 r5 w* Q7 |9 _" D* l
int op2 = ((Integer)(operandStack.pop())).intValue(); l, \, f' U" j5 X0 h7 \! h7 a/ z( d
operandStack.push(new Integer(op2 / op1));
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}0 p# Z) N7 y U/ B4 e' C8 B2 P% R4 [
5 y4 g$ B) _3 R; D( F) N class ImagePanel extends JPanel9 n5 J' g0 Y# L
{6 {% Q" E' @) k( i& w. x
public void sshow(): @' f% X% \; D# s* R
{. v* _; z' i( X; C* N- n/ _% X7 w+ Q
int i;
. n/ s# }( ~( ? for(i = 0;i < 4;i++)
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card[i] = (int)(1 + Math.random() * 52);# e# a& B: |( y. M6 b
}
& b, V5 V6 j7 R repaint();
" K, {2 I$ `6 F2 G. h1 K }
! B( ]! `; z0 D9 U# U& t/ }
- A4 u* p$ P# J2 v* s) u protected void paintComponent(Graphics g)
- l" z' U9 b y {
% o6 g. m& v8 ]3 M7 A9 q8 G; n super.paintComponent(g);" P1 O s+ T6 A' [
int i;' ?* G5 @8 o, ^8 c% g
int w = getWidth() / 4;2 `( l6 ?: d* d& l
int h = getHeight();
% M' m3 A$ {4 e* a0 }$ Y int x = 0;: g# }' \5 T$ A3 b9 W# e
int y = 0;6 g+ A( q- O) H3 G
for(i = 0;i < 4;i++)
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ImageIcon imageIcon = new ImageIcon("image/card/" + card[i] + ".png");
2 k# b9 l: C# W+ `0 o, L' Y Image image = imageIcon.getImage();5 k7 J$ v, ~) {: _4 n
if(image != null)7 k* `! w* I3 q
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g.drawImage(image,x,y,w,h,this);, T }: \, }+ h. u% B% R7 g& m+ g
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x += w; V2 D; A* l; Z
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}
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) x4 n; w) @5 R: Q$ N& L& H public static void main(String[] args)
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TwentyFourPoke_Game frame = new TwentyFourPoke_Game();
6 B" e* Q8 K+ l$ g) Z% s frame.setTitle("24 Poke Game");
0 v3 z8 O- e9 e6 [, L frame.setLocationRelativeTo(null);- F0 b6 B) a+ e7 Z3 E* @
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);0 ~* n3 U2 }: J8 D* `
frame.setSize(368,200);+ _% z0 Q2 Y( P" M5 t8 P
frame.setVisible(true);
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5 q) K% L% s, u' j# k } 复制代码 0 B* I; Z) s6 k: O
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