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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。
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( U, n% i5 L* e. A) t; c0 N4 Z 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。7 X1 a; C7 }( ?7 }9 l }
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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;, }! Y' [4 A: L. F! q- j
, A9 @, v b" Q4 I. Y$ {- R 以下是java 源码 :8 M) |; Q' L4 W: @! w+ O3 Y6 S
import java.awt.*;
, i/ v; R5 d- N import javax.swing.*;5 Z% J! ?) i' M* g/ C0 J
import java.awt.event.*;
; j: g7 J. ~0 u! O5 R! e* Q1 z import java.util.*;
9 o4 i4 Z2 Y6 ` import javax.swing.JOptionPane;
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# k( C- h. T! Y7 X: ], ] public class TwentyFourPoke_Game extends JFrame
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private JButton jbsolution = new JButton("Find a Solution");
1 L7 ?) n, t8 { private JButton jbrefresh = new JButton("Refresh");
+ t* j P; u( T3 ^9 a* A private JButton jbverify = new JButton("Verify");
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private JLabel jlmessage = new JLabel("Enter an expression:");' u3 e) p$ g# M) A( E `/ T( c
: b! s8 M$ o4 Z% o% V5 j% h private JTextField jtsolution = new JTextField();- Z2 k% h0 t& G5 |
private JTextField jtexpression = new JTextField();! {* _3 C, w4 [' t+ z- y( U' c
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private ImagePanel pp = new ImagePanel();
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private int[] card = new int[4];( P& }5 u( t2 X: i4 B
private double[] bcard = new double[4];
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8 L9 a+ ~7 _8 L3 X; p" _ private double sum;: F- w5 W! g& ^$ J! d! N
private double[] temp1 = new double[3];
b: F4 {, o& p! Z private double[] temp2 = new double[2];
; f4 W: _8 x' f# T1 M4 M; g6 e private char[] sign = {'+','-','*','/'};* a7 H# n; c4 i! @4 H2 r3 N( C. j
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7 b6 @; L1 Z& ]9 s" ` public TwentyFourPoke_Game()( Z+ Q, x2 G7 l' v
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JPanel p1 = new JPanel(new GridLayout(1,3));! y( G4 L7 I# a6 e; m/ \
p1.add(jbsolution);/ F7 [* r0 w% [) @# @' g4 {; R
p1.add(jtsolution);* x; {8 e, M* D, o5 r8 O
p1.add(jbrefresh);7 R7 y: x( w% Z4 J0 y7 w
JPanel p3 = new JPanel(new GridLayout(1,3));
& F, O+ H) x$ [$ L" }) D/ M& W" v p3.add(jlmessage);
2 G7 A* M4 G7 y# v1 h' j2 [ p3.add(jtexpression);- c7 |$ J' A" s; b" j( u
p3.add(jbverify);
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8 M7 K% I( D& h! Y, n3 e1 Z add(p1,BorderLayout.NORTH);
' {& ?: U# h+ d add(pp,BorderLayout.CENTER);
$ N+ L7 U; A2 u9 H( ^ add(p3,BorderLayout.SOUTH);
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ButtonListener listener = new ButtonListener();0 }7 c" J4 b% U B9 F% y
jbsolution.addActionListener(listener);
. y! C0 U6 d8 [( P, w+ z6 X jbrefresh.addActionListener(listener);8 B4 V' N! ?( D
jbverify.addActionListener(listener);
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class ButtonListener implements ActionListener
4 l. ]6 U3 i+ O8 e$ { u {
; K5 d0 _7 Y8 J9 @5 v public void actionPerformed(ActionEvent e). s5 F0 C4 y' t8 d7 b
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if(e.getSource() == jbsolution)
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* X( F8 Q4 n6 X$ d" | for(int i = 0;i < 4;i++)7 T0 U! [' P$ g
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bcard[i] = (double)card[i] % 13;1 x+ S) N- Q R% e! _* q: {
if(card[i] % 13 == 0)6 V' {# c% U1 J
bcard[i] = 13;
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search();! m, a0 z' p' j2 V% z' ]
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else if(e.getSource() == jbrefresh)
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pp.sshow();
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9 z+ E) e( ?8 q* Y else if(e.getSource() == jbverify)! Y4 s. M x8 T. I# n t
{
" g, x1 G2 _: A8 Q- K2 q/ T String expression = jtexpression.getText();& w2 A( T( |% b* W' G& g
int result = evaluateExpression(expression);
3 K- z. x7 e' U if(result == 24)
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JOptionPane.showMessageDialog(null,"恭喜你!你答对了!","消息框",JOptionPane.INFORMATION_MESSAGE);
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else
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. @) P& W" R4 j9 o JOptionPane.showMessageDialog(null,"抱歉!请再次尝试。","消息框",JOptionPane.INFORMATION_MESSAGE);* H: _9 V1 h1 a: [
}
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8 n0 Z9 m- i' U( n7 S public static double calcute(double a,double b,char c)
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if(c == '+')
$ U' w; j! }4 z( A return a+b;
5 T6 g6 j& ~6 \" `# Q6 L, Z: a1 d. O else if(c == '-')
8 Q8 M: S" e! L8 T+ L9 C: O return a-b;2 U) [3 \: v, C2 m2 J! T+ q' o7 b
else if(c == '*')
* t K, V8 d5 @- O& w return a*b;
6 W& \8 X$ t9 m. ^" x else if(c == '/' && b != 0)% y6 h+ c a- {- X
return a/b;: I- ]$ z, z2 j9 l/ M. O
else% t/ G) G0 M" K) \
return -1;
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public void search()) b4 D1 H7 |' @9 [, D! l: B
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boolean judge = false; |; F6 p6 j0 Z
for(int i=0;i<4;i++)2 a$ b! ]9 s& O. f+ d" v$ l9 n
//第一次放置的符号
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for(int j=0;j<4;j++)
9 L, G9 c5 D \ //第二次放置的符号0 |! x7 G& W4 l& P' ]
{
: `( I. y: s7 w" a( C# g7 N" K+ \ l for(int k=0;k<4;k++)! U/ }: }/ L1 N
//第三次放置的符号4 z# t6 b! v* n" u
{
8 g0 E5 R. r+ g, ~ for(int m=0;m<3;m++)
# A: O; y9 @: s' R4 v X //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
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if(bcard[m+1]==0 && sign[i]=='/') break;
`' B' v; B. f+ \, A. T( h% R temp1[m]=calcute(bcard[m],bcard[m+1],sign[i]);
$ D) S2 d* @- ]- k* }! Q8 w0 [& Y. Y temp1[(m+1)%3]=bcard[(m+2)%4];
, v8 B) o, A4 |" O temp1[(m+2)%3]=bcard[(m+3)%4];6 t% W0 W0 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
% Z; A3 ?0 I* h# W for(int n=0;n<2;n++); _+ J; E5 K/ ~
//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7 l3 C3 L: U$ Q3 r6 A3 r* v( t, U! p
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if(temp1[n+1]==0 && sign[j]=='/') break;' Q) ]* i* d5 u* v1 U
temp2[n]=calcute(temp1[n],temp1[n+1],sign[j]);
7 Y4 H5 G8 S# P& K" p1 K) G8 m* [ temp2[(n+1)%2]=temp1[(n+2)%3];
5 T) @% m5 V9 Q/ E //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
. R* D7 P7 V c0 A0 k if(temp2[1]==0 && sign[k]=='/') break;
5 @ Q) ]6 l- Z- B sum=calcute(temp2[0],temp2[1],sign[k]);
3 D3 H: I& e$ _! b# E7 M //计算和
8 e- N+ E2 t& t1 O1 a if(sum==24)3 E/ I" R: e0 @) G7 W1 A
//若和为24
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judge=true;
h/ J: L# a ]( J3 b+ ? //判断符为1,表示已求得解
' U. _, f/ [/ d, W, X2 K( ? if(m==0 && n==0)( _/ ?7 s% b% G" ?$ ^0 t
{
# u% D2 @9 M5 h: X2 N; E String sss ="(("+(int)bcard[0]+sign[i]+(int)bcard[1]+")"+sign[j]+(int)bcard[2]+")"+sign[k]+(int)bcard[3]+"="+(int)sum;' \3 u; m' v, z7 q6 H( p) ]% D
jtsolution.setText(sss);+ D. c6 A/ x/ Q1 b# ~9 Z5 }; a L
return ;
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else if(m==0 && n==1)
" D$ d4 C4 Q+ U, E3 K+ Q {
9 A4 K' H, O: W O& V0 w0 ], q& x String sss ="("+(int)bcard[0]+sign[i]+(int)bcard[1]+")"+sign[k]+"("+(int)bcard[2]+sign[j]+(int)bcard[3]+")="+(int)sum;+ C! c- r1 n6 b. I& P2 W, N
jtsolution.setText(sss);
" K3 V0 k1 `. K8 q% j/ [! K return ;
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else if(m==1 && n==0)
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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;
$ r# m, [" C; x jtsolution.setText(sss);9 o6 Y2 x! V9 Y
return ;/ U5 }8 h4 t3 S* r8 j1 K
}
; U+ g7 U" e5 V( ]* r3 |5 V else if(m==2 && n==0)! V" G3 O! V# K% Q- W! M3 m' V* H
{
9 y4 d3 K3 f8 l8 v" L7 o) P7 h String sss ="("+(int)bcard[0]+sign[j]+(int)bcard[1]+")"+sign[k]+"("+(int)bcard[2]+sign[i]+(int)bcard[3]+")="+(int)sum;
. ~: ]! a- B4 n4 X* n) Y8 K jtsolution.setText(sss); J2 d; k' n4 ~) u5 O
return ;" Y) x+ W. `8 }. \
}
$ n7 x1 N% l7 q4 B. d1 b+ B else if(m==2 && n==0)
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String sss =(int)bcard[0]+sign[k]+"("+(int)bcard[1]+sign[j]+"("+(int)bcard[2]+sign[i]+(int)bcard[3]+"))="+(int)sum;
- Y! C6 F; w: H8 w6 |( I7 H9 C jtsolution.setText(sss);2 F* V' R. T' \, D' I+ J% n3 n
return ;
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//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, O# m7 \: N2 E; m: Y: J1 h3 g& i8 F- n
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}
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}
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3 m7 M C0 y4 O7 u T4 K H! c if(judge==false)6 ^" V: a( ?3 _, a4 I, R, q
jtsolution.setText("No solution!");
3 W, D; S/ p, U. N //如果没有找到结果,符号位为0
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7 @7 ]" R) F2 Q2 O- J+ s; X public static int evaluateExpression(String expression)& c' _3 B. j2 N
{
4 U/ z; U A, t. l% S/ k // Create operandStack to store operands( p6 ^, D7 Y( t7 g8 p1 I* ?
java.util.Stack operandStack = new java.util.Stack();
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// Create operatorStack to store operators
2 ?& c$ M1 o5 S; \; ]$ Z% @ java.util.Stack operatorStack = new java.util.Stack();
6 C- C% M( C) R8 `
2 z a P& S$ H, T; V6 ~6 ` // Extract operands and operators
2 _) L# ~/ L. ~9 _ java.util.StringTokenizer tokens = new java.util.StringTokenizer(expression, "()+-/*", true);/ W1 z* @( R2 E7 D, U
2 q! [5 l* T! l8 l" T // Phase 1: Scan tokens
, x/ Z6 S$ G+ Y! v while (tokens.hasMoreTokens())
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String token = tokens.nextToken().trim(); // Extract a token
; z2 ~6 K- w. j4 m! Q if (token.length() == 0) // Blank space5 t% _- l5 o( R
continue; // Back to the while loop to extract the next token. n& } G9 K2 u# @, x( f6 v
else if (token.charAt(0) == '+' || token.charAt(0) == '-')
" H, v2 g, }. N, @" p6 z6 t/ a {
; a' k( D6 k9 i% D" a L# ?3 F // Process all +, -, *, / in the top of the operator stack7 Z$ y& q0 i. q8 X6 \
while (!operatorStack.isEmpty() &&(operatorStack.peek().equals('+') ||operatorStack.peek().equals('-') || operatorStack.peek().equals('*') ||
' c G+ `" m) V operatorStack.peek().equals('/')))
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processAnOperator(operandStack, operatorStack);/ T. X: L; Q1 G- e$ @. T& F& _2 i
}
{8 Z" Y/ t7 U9 v! m: } // Push the + or - operator into the operator stack
! m2 w l$ Z' l operatorStack.push(new Character(token.charAt(0)));6 e. N2 ?% P' {% P! k
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else if (token.charAt(0) == '*' || token.charAt(0) == '/')8 Y8 W, ?- { F2 x! Y1 m3 k5 t/ w
{
2 d4 ^( @6 u& g9 w( w( ]+ c) ` // Process all *, / in the top of the operator stack
$ x3 H# r2 _$ K+ k2 a- X while (!operatorStack.isEmpty() && (operatorStack.peek().equals('*') || operatorStack.peek().equals('/')))0 M! f' N0 w3 [6 N9 _7 t8 P; n
{8 [9 J7 a- j. X' J4 s' Y( H
processAnOperator(operandStack, operatorStack);# ?& H9 q9 a% O
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// Push the * or / operator into the operator stack
5 d. n c3 V. a. ?4 t+ e operatorStack.push(new Character(token.charAt(0)));* J7 z6 S8 I& E; ~
}
) D/ q9 C* K# {9 Y- g else if (token.trim().charAt(0) == '(')
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operatorStack.push(new Character('(')); // Push '(' to stack+ o2 Z' |% c8 {6 K B1 M" f- B) |
}
6 H# U! b6 w" o else if (token.trim().charAt(0) == ')')
! l3 e) o8 z! p {, f- a d. x3 R2 p$ i
// Process all the operators in the stack until seeing '('
/ ]6 ~0 h5 _8 _ P, ~: n while (!operatorStack.peek().equals('('))% C9 d0 i' n; c1 D% _" g
{
7 F$ |( J# r' K) ]7 h4 B, C processAnOperator(operandStack, operatorStack);
, ^- d& w d2 g& l0 q% K8 } }
- l0 f9 N0 y. B6 E* H4 Q operatorStack.pop(); // Pop the '(' symbol from the stack. h. K/ }2 e& p4 h' |
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else* R& @9 B* I! s) L+ m& r
{: I0 b; i- Q$ Q/ e$ n% a9 W* }
// An operand scanned" @- x6 p8 u h" O% |
// Push an operand to the stack
! S# j3 Y9 S4 b5 p& [- _" D) _ operandStack.push(new Integer(token));+ X8 K; ~- J% d; y. \- C
}- W( T' F9 j8 B+ r3 I+ L
}
# G& X9 l. H9 [/ z3 e
* ~: ?/ G7 H B9 Y' x // Phase 2: process all the remaining operators in the stack
; o9 G$ D6 q& I" Z: R while (!operatorStack.isEmpty())* {% j L6 n- g. `% U8 u
{" u8 O0 R& b; T: |2 G! F% k* {
processAnOperator(operandStack, operatorStack);
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, x7 h2 x5 o: ^, C4 X3 t // Return the result9 d& B# o& N5 l4 X% S5 W1 ~9 i
return ((Integer)(operandStack.pop())).intValue();
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: |5 v) s; M5 y% L: F& w/ W. z w5 P8 J3 c* l# v ?; i) k
public static void processAnOperator(java.util.Stack operandStack,java.util.Stack operatorStack)0 I# A7 [$ T7 e% R! u" F% v8 h
{
7 |+ f/ o9 `& |7 C if (operatorStack.peek().equals('+'))9 f! J1 d( U2 r6 e
{
3 A! l, R$ y, }! E operatorStack.pop();$ u. E2 P& Z2 H. X8 ~
int op1 = ((Integer)(operandStack.pop())).intValue();
# A) d/ K. H- w! ?6 l" N0 J int op2 = ((Integer)(operandStack.pop())).intValue();1 ~" G5 `; U# X& Z
operandStack.push(new Integer(op2 + op1));
" A7 n* r+ X& R+ r7 A: t9 q }
4 K6 ^* ?. t, o% `; m1 j0 m else if (operatorStack.peek().equals('-'))
% e! t/ |+ I& G; E4 x# H {
' v! r' i" n. w operatorStack.pop();
% v9 { x* y% _9 D& e( |9 t int op1 = ((Integer)(operandStack.pop())).intValue();4 `5 @7 g. s# _
int op2 = ((Integer)(operandStack.pop())).intValue();+ ], G2 C. k8 @% _
operandStack.push(new Integer(op2 - op1));
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else if (operatorStack.peek().equals('*')), h7 F3 L4 j+ Z( b
{
( J6 u3 I$ G6 }5 U# E5 T% L operatorStack.pop();
8 E! p3 T( E/ t1 T! B/ R int op1 = ((Integer)(operandStack.pop())).intValue();
3 `% P! @7 e2 a int op2 = ((Integer)(operandStack.pop())).intValue();
* _( e: I' K6 w3 P( v6 c operandStack.push(new Integer(op2 * op1));
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else if (operatorStack.peek().equals('/'))
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operatorStack.pop();
' H% g; ^. `8 g l8 }* f1 q7 s3 F int op1 = ((Integer)(operandStack.pop())).intValue();
% ^0 f8 P+ K" U! }% Q+ r int op2 = ((Integer)(operandStack.pop())).intValue();9 r/ r3 Y+ t$ o9 Z
operandStack.push(new Integer(op2 / op1));# L4 U C# m. C* v6 [' U5 M4 q
}
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% B; q0 Q% J- @8 H class ImagePanel extends JPanel
4 ~# Y. Y* K$ s' f6 h# E; O {
7 I$ z, M% f* ]* d" d/ F" ? public void sshow()
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int i;' N2 O% C- P% w0 j/ `0 i- F
for(i = 0;i < 4;i++)& B, U- b* T* }* F* Y, _
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card[i] = (int)(1 + Math.random() * 52);
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) r, X9 y9 K" A* h$ ~ repaint();. _1 g6 C$ Q+ R7 D# i: A+ m
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x/ S2 ]; D D1 k+ t protected void paintComponent(Graphics g)* r; K2 A% P2 M+ C" ?' w( [
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super.paintComponent(g);8 ?$ \4 B# ?' P
int i;# j0 X/ }/ P3 w% k
int w = getWidth() / 4;
1 @" I0 T7 H: a N0 Y int h = getHeight();
: s0 I' P" s8 R/ d$ ~5 v* b) ? int x = 0;' t6 P/ C+ t9 P& o: e, _# K! P& r* w3 K
int y = 0;6 N9 \* x2 _; J7 r5 L
for(i = 0;i < 4;i++)
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ImageIcon imageIcon = new ImageIcon("image/card/" + card[i] + ".png");
3 C4 z, `3 t+ c t! S* }) x Image image = imageIcon.getImage();: ^: i8 Z5 i& _! F4 ?0 {
if(image != null)
' J2 l* X) A$ k5 j {
( d6 [% ]; e2 M% X g.drawImage(image,x,y,w,h,this);- q* [( W* S& n$ v- s# Q! i
}
3 X5 w. K, q' c. j! L( ` x += w;
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}; U5 _+ r! N7 U4 W$ G1 B
}
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public static void main(String[] args)& Y* e- G) `) k6 l2 v/ ~
{
p$ l- y2 w) A+ o1 g- q TwentyFourPoke_Game frame = new TwentyFourPoke_Game();+ Q- _6 r) f1 U
frame.setTitle("24 Poke Game");: f7 w0 [- [8 [' z) ?. s
frame.setLocationRelativeTo(null);' w L: g, d ^* j0 d9 M
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);, h3 J E/ H8 @* m
frame.setSize(368,200);1 P5 h$ F' _8 m" \7 P5 h
frame.setVisible(true);
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1 e( g! l/ a) { } 复制代码 % y7 X u" E1 I7 U' i6 N' }
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+ I( w) e( K! S% R3 v; {. ]
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