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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。# F" a0 v/ f- b$ Q, M2 \
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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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* L& h! b. e' U! b5 F 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。
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3 T9 m( x: k: g D; A! Y" {. z 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;. a! N& R1 s) r" b
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以下是java 源码 :4 m$ [( H, P: c b0 y9 X8 a
import java.awt.*;
0 ^- N4 J3 s2 F) \+ L) n import javax.swing.*;, \9 D+ y* e3 z" W1 e% m
import java.awt.event.*;
% `, N$ t0 }/ ~ import java.util.*;
( E/ \+ s' X' `, u4 {5 Y1 e8 j. d import javax.swing.JOptionPane;
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public class TwentyFourPoke_Game extends JFrame: j$ _' p# Z* m4 T7 I- M4 r- P, F
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private JButton jbsolution = new JButton("Find a Solution");
4 j3 z! G) l. X9 O private JButton jbrefresh = new JButton("Refresh");% ^8 P) T8 _0 s$ w+ ^" I
private JButton jbverify = new JButton("Verify");# i; {- a# X2 j, i7 b+ m3 A& l
; P/ b& w8 X% ?! l
private JLabel jlmessage = new JLabel("Enter an expression:");* v* _* u; N1 Y. P
0 r* Y" r7 m( A; v% T private JTextField jtsolution = new JTextField();
4 h8 @4 e5 G, D private JTextField jtexpression = new JTextField();& S! g- \# P4 L/ g6 K
% u8 T7 z) ^2 E8 Z5 _$ V4 Q3 Y- B private ImagePanel pp = new ImagePanel();
' W2 b" U" S! v
6 \$ X, [$ g* b" Z private int[] card = new int[4];. @. [9 k8 c+ @0 Y0 _
private double[] bcard = new double[4];. l" Z% c- n+ l; m
2 j# Y$ N7 q, H) r
private double sum;4 a9 p/ t# [% l$ k: V: U
private double[] temp1 = new double[3];& @0 h4 P- J( v5 a+ z r( ]
private double[] temp2 = new double[2];
" R+ x, Z& r8 D6 a6 m private char[] sign = {'+','-','*','/'};8 o1 c% [/ |- @+ d' i
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public TwentyFourPoke_Game(); ~9 ]1 m# k! u8 S
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JPanel p1 = new JPanel(new GridLayout(1,3));
7 _+ C+ k% ^5 e5 s- {" y/ a p1.add(jbsolution);: c' l! H2 \4 |
p1.add(jtsolution);
2 @' t# h/ l5 x p1.add(jbrefresh);# u5 m0 N8 W# \( E
JPanel p3 = new JPanel(new GridLayout(1,3));
2 p2 r! Z$ r; X' s# Q+ W* P9 G/ `! z7 i p3.add(jlmessage);3 [: v3 T# t6 q- q
p3.add(jtexpression);
8 |' V; a: y0 w/ G% p9 o% l0 g p3.add(jbverify);# v' e3 N" q/ k/ `: O$ }
* J, G% K! i; q. F/ v& U6 E( w add(p1,BorderLayout.NORTH);9 J* t. x% \2 k2 E2 M
add(pp,BorderLayout.CENTER);4 ~3 {& E# L( x
add(p3,BorderLayout.SOUTH);: f5 f/ A2 D( p* s* \3 _9 v% w
& z. g4 Q) h# v+ |7 o9 x" O @
ButtonListener listener = new ButtonListener();! t& [# J+ m; e8 W m- K+ t
jbsolution.addActionListener(listener);
' [& Y9 N" W+ b( _; \( X jbrefresh.addActionListener(listener);8 G! y5 O; z2 S8 d, ^
jbverify.addActionListener(listener);8 ]/ G! j7 _+ P7 e0 s8 v
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8 I& g/ h7 P/ N" [+ p7 T3 _' J/ [, ^
class ButtonListener implements ActionListener1 ]7 L- m: I" s# G& }& M
{$ q e" W2 I4 j
public void actionPerformed(ActionEvent e)$ D& ~- i+ Y' a
{9 B3 t7 M8 Y" j$ T0 D; a$ S
if(e.getSource() == jbsolution)
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for(int i = 0;i < 4;i++)2 i- z# n9 Q1 x% J( @2 ?
{+ Q. m/ r5 X: l- j( Z9 y
bcard[i] = (double)card[i] % 13;
6 _7 v( |3 _6 }5 C) T: H if(card[i] % 13 == 0)7 J/ V, M/ t6 q! b3 u" \) G; a U
bcard[i] = 13;
/ y" f7 d8 Y% i3 R3 y+ D1 w }; D; Y, Q, p; z: _ k
search();/ `* B, Y% i! C9 y+ c. R- V/ w
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else if(e.getSource() == jbrefresh)
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4 `- w" I Y5 P+ T* H, R9 b pp.sshow();$ O) v/ }" y1 h) k( B+ k. H
1 s& I% n8 l) X- ^; z }
; {4 [0 m# u* v. j' w- C else if(e.getSource() == jbverify)
" A% Z. E1 p" o; i$ I1 j! S {
3 }# P% `( \: n String expression = jtexpression.getText();
6 E ~. ^- m$ R/ L: g" U% [& g, ]* W int result = evaluateExpression(expression); q- j3 y+ s, l$ b
if(result == 24)0 O5 Q5 ]5 Z$ M i
{
7 ^" _3 |4 X' D3 C& o JOptionPane.showMessageDialog(null,"恭喜你!你答对了!","消息框",JOptionPane.INFORMATION_MESSAGE);! V3 N5 N" `0 k% I
}! M4 k0 V& ?# L6 f- Z
else6 ?9 ^. O; q% o" t$ S6 H R3 I
{
w4 f' `0 _# z v JOptionPane.showMessageDialog(null,"抱歉!请再次尝试。","消息框",JOptionPane.INFORMATION_MESSAGE);
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8 V# k) w& v. ~' f& m! h public static double calcute(double a,double b,char c)
/ j8 c* x8 v" ~4 b# \3 g {
$ }0 a* @# A& j/ g D1 r if(c == '+')0 }! i8 R2 W; d0 S( T
return a+b;" I, [. ?2 J$ b( t/ \
else if(c == '-') R( _ `6 Y; V% a
return a-b;
" ^# o6 U& W4 B4 h7 U else if(c == '*')/ l# B6 W" O. m& O" z* U
return a*b;
$ P) c4 ^) Y5 J1 x else if(c == '/' && b != 0)( N) B. P) h5 |! H3 u# X* m
return a/b;
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return -1; L0 O+ }. Q% a
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public void search()
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boolean judge = false;4 n( E$ Y2 Q+ Z2 D
for(int i=0;i<4;i++)
+ u/ P$ P$ _4 }( Y //第一次放置的符号
) q# \) {# E& x- b' g4 F$ j2 P {
8 \0 L- ~& S/ c8 A) c for(int j=0;j<4;j++)9 T; ?) x1 y4 d8 Q' F3 [1 w
//第二次放置的符号
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for(int k=0;k<4;k++)
7 b& x4 F+ Q( d+ |( b //第三次放置的符号
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for(int m=0;m<3;m++)
: S1 x: E3 h0 ~ //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
* Q; U% N9 b# U3 M) N {
5 h2 ^: A! Q, \) | if(bcard[m+1]==0 && sign[i]=='/') break;
2 c; y7 i! Y6 R/ m$ I$ B' H temp1[m]=calcute(bcard[m],bcard[m+1],sign[i]);
- N& J1 ]/ E3 ]/ f* e3 v0 g7 e temp1[(m+1)%3]=bcard[(m+2)%4];
' T3 C+ p: H0 I' _( e8 g% p3 S. i: k temp1[(m+2)%3]=bcard[(m+3)%4];
& |5 p8 e9 ?; z* d2 [' A //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' p+ b2 e8 x/ n: T% }9 ^0 Y# h
for(int n=0;n<2;n++)# h+ v7 J9 q7 Z- \
//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% Z; p8 W1 x5 ^% {& J
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if(temp1[n+1]==0 && sign[j]=='/') break;
* x$ J$ |0 M5 I% t5 {; s temp2[n]=calcute(temp1[n],temp1[n+1],sign[j]);
2 u) r) _' Y3 j temp2[(n+1)%2]=temp1[(n+2)%3];
, ?2 K2 ~! M' @% h d //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
: W2 }7 N$ k! `8 k2 ~( m, \ if(temp2[1]==0 && sign[k]=='/') break;
9 |$ q5 x7 o3 ?+ d sum=calcute(temp2[0],temp2[1],sign[k]);
" b" H# @4 u1 w. Y4 P9 {; x3 D5 ` //计算和
+ v& D( r" W: X7 @- j if(sum==24)( G# O7 J. J1 z2 S! ^ L0 x
//若和为24
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judge=true;
! ^5 m/ |- `! T( R //判断符为1,表示已求得解& @8 o; {: U% F
if(m==0 && n==0)4 Z8 y6 r0 j0 ?
{
- w7 Z' R2 x, K$ N5 X8 U: L String sss ="(("+(int)bcard[0]+sign[i]+(int)bcard[1]+")"+sign[j]+(int)bcard[2]+")"+sign[k]+(int)bcard[3]+"="+(int)sum;1 }5 ?/ l9 D9 l7 d) E' e4 \$ @
jtsolution.setText(sss);2 M. S6 s7 R3 w# [7 h
return ;
2 j+ O5 M# M/ u- N6 }: U& c }
* q+ S% H1 V1 b% s b5 V2 C else if(m==0 && n==1)" X1 M+ D2 z4 R0 a) ]- O3 o1 }+ A
{
- b* f1 A4 C5 ?( D String sss ="("+(int)bcard[0]+sign[i]+(int)bcard[1]+")"+sign[k]+"("+(int)bcard[2]+sign[j]+(int)bcard[3]+")="+(int)sum;3 S/ g4 C# A8 r( `! C9 t/ M
jtsolution.setText(sss);2 L5 }; V5 C+ W. i
return ; R& {9 H3 S. u# B/ `
}
5 p: s* [) |- h) j+ ~+ h) j d else if(m==1 && n==0)1 E/ w# C- q" @2 l* e
{
; U% Q7 q; I7 l String sss ="("+(int)bcard[0]+sign[j]+"("+(int)bcard[1]+sign[i]+(int)bcard[2]+"))"+sign[k]+(int)bcard[3]+"="+(int)sum;
7 G* e; S3 R: d' v3 G3 c jtsolution.setText(sss);( O, O2 _- m3 ?1 c9 k- k. F0 W6 J
return ;. e& Q# B% l* t% Q. S
}
9 t8 j5 _$ V2 w0 C% y: n else if(m==2 && n==0)$ p$ }. Z6 f" y- w& c$ ^
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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;
+ [- I. M( @2 W; n9 P9 ` jtsolution.setText(sss);8 t6 Q6 W/ ^; w7 [
return ;
# _7 `2 f& g/ \0 h }
+ k' ^( m' ?4 m. A) B else if(m==2 && n==0)* s- z/ P! D9 B
{
1 J# ]: H) E5 p) ? String sss =(int)bcard[0]+sign[k]+"("+(int)bcard[1]+sign[j]+"("+(int)bcard[2]+sign[i]+(int)bcard[3]+"))="+(int)sum;) b5 D) s0 p6 j/ T6 Y
jtsolution.setText(sss);- I/ I$ A' U) f8 Z( D6 v
return ;2 `& Z- X+ l/ a! K2 ]
}
' Q4 P; R4 a8 S' U M) A3 a //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
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C* q7 }6 V2 T7 n) Q+ P }
1 Y7 w# r3 \" L/ E) S d0 q }
5 {( |4 M/ ?. [- E8 G% y }1 @9 G0 z+ {( Y8 F @6 u
}
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}
+ r P* a9 }4 H if(judge==false)
1 H/ p X: C3 o: m2 A- O( b jtsolution.setText("No solution!");
4 t1 [, F, q: ^3 ` //如果没有找到结果,符号位为0
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public static int evaluateExpression(String expression), B2 ?! y1 V& e& K% ~* @" C
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// Create operandStack to store operands6 F+ o7 Z0 V0 r L
java.util.Stack operandStack = new java.util.Stack();# \' D* V3 o0 t9 v& [; C2 R5 z
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// Create operatorStack to store operators! |3 N6 ~3 j! L+ [! u3 L
java.util.Stack operatorStack = new java.util.Stack();
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# u, a t8 K3 o8 t$ {/ t // Extract operands and operators
) z7 a! R+ G2 U java.util.StringTokenizer tokens = new java.util.StringTokenizer(expression, "()+-/*", true);
- x4 H0 R1 L1 \# Y1 l6 J' H) x
3 h+ i3 Z( j% ^+ ?) l: q // Phase 1: Scan tokens J+ A d e; S
while (tokens.hasMoreTokens())
6 w4 \# n7 R: k( h {
Z) y! t6 {' ^; o String token = tokens.nextToken().trim(); // Extract a token# ~+ e0 e, c% S: f; G- o! l+ A
if (token.length() == 0) // Blank space
6 j1 T1 r1 k( c continue; // Back to the while loop to extract the next token
9 T: o0 @- C/ C% ^6 u' s else if (token.charAt(0) == '+' || token.charAt(0) == '-')
5 j' `% v" G( _' b- j; h {
2 j0 i6 b u. c) B9 M- C // Process all +, -, *, / in the top of the operator stack
1 E3 D) K' S* G' {# Z8 l/ J while (!operatorStack.isEmpty() &&(operatorStack.peek().equals('+') ||operatorStack.peek().equals('-') || operatorStack.peek().equals('*') ||7 w5 Z" T$ k' q8 X; g: w ]( c
operatorStack.peek().equals('/')))
2 R/ y& U8 m' |& L1 o" V" ?! I$ i {
% m6 r6 Q& s3 u processAnOperator(operandStack, operatorStack);
% x3 \. ~5 \- m4 F2 n5 J }
( Z4 ?7 r, T ^; N: d7 z4 q // Push the + or - operator into the operator stack
) u/ M: u X8 d6 j1 a/ F operatorStack.push(new Character(token.charAt(0)));$ @% `+ U: j* B: U. x1 H
}
) E5 ^6 p( l. h. d5 a! b else if (token.charAt(0) == '*' || token.charAt(0) == '/'). E0 e) Y0 V5 w0 @0 h
{
. s6 W$ ~8 d8 d9 f& e N# Q" a // Process all *, / in the top of the operator stack
% {- _3 P% M, ^/ `+ l& k4 ]% A while (!operatorStack.isEmpty() && (operatorStack.peek().equals('*') || operatorStack.peek().equals('/')))/ I: E+ p* i# J
{( f: J7 ?9 x7 y6 r( ]
processAnOperator(operandStack, operatorStack);) r7 W \. ]2 G
}
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// Push the * or / operator into the operator stack& s3 X0 ^* d" T
operatorStack.push(new Character(token.charAt(0)));
' s6 @& q0 U, C- f' L; { }
3 V% H, m4 h! c7 H* G5 q else if (token.trim().charAt(0) == '(') p6 J+ y# J9 O% Y# X
{
9 X& v+ T) Q6 W. Y7 j7 [ operatorStack.push(new Character('(')); // Push '(' to stack, G K: l' w& U) c& ^ g
}
8 r7 ^# |7 I: z) U7 } else if (token.trim().charAt(0) == ')')
, R; t5 i- Y; @( Z. p3 R" u3 D {
3 H8 o3 n I; `* o // Process all the operators in the stack until seeing '('
5 E" ^" m0 [& v* f4 J while (!operatorStack.peek().equals('('))
/ @% S" R% |( l( V6 u {
! U6 j( b# U$ L8 { processAnOperator(operandStack, operatorStack);: d8 a9 F7 m! ]" R; R! E
}
/ V0 y8 i. y, U9 E operatorStack.pop(); // Pop the '(' symbol from the stack1 h. o1 O. Q/ A- O4 g& v( E
}: [5 s q* `/ S z. \
else
5 d/ F* ]8 K2 K# m" T2 `) [ {
$ P$ n; x+ D8 W# J& q2 E& _- m // An operand scanned
5 _/ {+ k( z# }, o1 [8 u // Push an operand to the stack3 b, g. o5 Y$ |3 D3 ~
operandStack.push(new Integer(token));: Q% P% m, V, |$ j; [
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}3 X4 q+ T# x/ v9 `0 h
# R' @0 m/ ?$ S/ ` // Phase 2: process all the remaining operators in the stack
6 X8 m: i0 Y8 l. T4 s3 J while (!operatorStack.isEmpty())$ X; _6 q" ?) h4 u0 w
{# \6 {3 k! h4 H
processAnOperator(operandStack, operatorStack);
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// Return the result
# i6 B1 R+ w( c5 S3 n+ r2 W7 `1 ^9 ]; | return ((Integer)(operandStack.pop())).intValue();
$ h+ z& s) g5 k9 w }# x2 a4 P7 d+ b4 R
: D/ X$ e7 X- @5 ~ public static void processAnOperator(java.util.Stack operandStack,java.util.Stack operatorStack)
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if (operatorStack.peek().equals('+')). J ]0 H8 H. l( F) R7 m' t
{
, q( D3 [$ K8 l T2 W operatorStack.pop();
' b' `1 u8 g+ m9 W% D; c int op1 = ((Integer)(operandStack.pop())).intValue();2 w. ]2 U" ~& ?
int op2 = ((Integer)(operandStack.pop())).intValue();0 t! r+ T, o5 C, \9 G5 e/ k- v
operandStack.push(new Integer(op2 + op1));5 x' t5 U$ _: G' u; {( c
} I% {; z# j$ h
else if (operatorStack.peek().equals('-'))" a/ E8 ? ]/ P: r- m- P0 H
{, p1 A, T9 b" O0 ?' a) B
operatorStack.pop();" C ~6 w! d/ V' n( Y: ~* L
int op1 = ((Integer)(operandStack.pop())).intValue();( x9 B) d! K! p+ h3 p
int op2 = ((Integer)(operandStack.pop())).intValue();
; ^# _, h3 e) O& g) o% L( ^ operandStack.push(new Integer(op2 - op1));
# J% ?. n* w0 l3 `, i' g' \4 \/ ] }
5 V- u3 D1 e# T7 j& |1 C- y! b else if (operatorStack.peek().equals('*'))
/ d" W6 }& |) t% R" H/ _& ^ {
" V( P* @( q# R operatorStack.pop();
5 p0 N9 F! C9 @' }- D8 @ int op1 = ((Integer)(operandStack.pop())).intValue();
6 S" s, Z# O+ P3 H7 P; n int op2 = ((Integer)(operandStack.pop())).intValue();
: v Y- ?! p% W% L operandStack.push(new Integer(op2 * op1));
6 z9 S3 d0 s4 A1 g' y: [ J7 ~+ h0 I! h }$ s1 j/ {) T6 s' ?
else if (operatorStack.peek().equals('/'))
, s$ j# C t [: V- Z {
5 h% S5 Y! D, Y! L operatorStack.pop();
; f1 ~% N& s3 `- @/ s5 }0 r int op1 = ((Integer)(operandStack.pop())).intValue();# J; u t" b6 M) H' R7 \& h5 n0 t
int op2 = ((Integer)(operandStack.pop())).intValue();
: [9 J% a% i, V operandStack.push(new Integer(op2 / op1));7 l4 M, ]1 Z7 C8 ^$ w Y3 o
}
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$ `3 k/ Z' J) H7 {7 M8 n5 V 6 E$ ~9 i" K/ ]
class ImagePanel extends JPanel
$ X/ Z& w& b z {( `' h- u/ G( A3 r* F
public void sshow()2 F2 A: h( k6 V, H- n9 X) b; p9 Q. d$ a
{- o, [; n) C8 \9 P4 t# e1 L
int i;
0 ?9 @- Q" S1 f8 @ for(i = 0;i < 4;i++). ]" X- p: G( W. R {9 c0 Y1 ?) ^: K
{) _( A5 F0 ?/ z. J$ y$ N5 _/ G0 l$ c
card[i] = (int)(1 + Math.random() * 52);
; h+ k, {+ E$ Z9 Z. ^ }
& Z% @& ^: Q) q repaint();: D7 `' ~- ~! O& W) O( Q/ `! ~
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. F5 h; h) F5 ~ protected void paintComponent(Graphics g)
- i" |8 ?; P" y8 Y9 n. I {
( R9 I3 R6 F' X5 m& @- N super.paintComponent(g);
/ E1 \# K& T% v4 }4 ~ int i;- a2 n# L: ]+ [; {) b
int w = getWidth() / 4;0 B% w: ~/ E E& r; _" F
int h = getHeight(); z+ ?$ h5 D @) |/ t7 ^" N, |. F% t
int x = 0;
- T9 s9 @# @* \6 ? @* v int y = 0;
# ^5 \& J/ h9 _+ F K8 r for(i = 0;i < 4;i++)
6 w8 O7 p6 Z% W {
9 p) S: S( h5 {4 N ImageIcon imageIcon = new ImageIcon("image/card/" + card[i] + ".png");
( G: w4 n. t5 @, t0 d R: H8 a, b Image image = imageIcon.getImage();% j. f f( U" E5 R
if(image != null)
( F2 o9 ]! G4 I: g {5 e4 Q. v* h4 P
g.drawImage(image,x,y,w,h,this);4 ~" N0 j! ], ~1 `# ~
}( K" Q$ |# A9 [, v. s
x += w;
$ f* n2 p1 `, v7 q }
}3 v5 |6 u+ U2 n4 o% [ }0 x% {* X- V' s( F- E }7 s
}
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public static void main(String[] args)& ?6 i! S1 k; }! s, j8 |
{
4 X4 o, G K9 @/ o& J TwentyFourPoke_Game frame = new TwentyFourPoke_Game();: {. o7 J( }1 l" g; H( U
frame.setTitle("24 Poke Game");' H& Z' {* L" t
frame.setLocationRelativeTo(null);
0 w8 p9 o) I2 ^ frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);8 j5 H/ }* w K( Y
frame.setSize(368,200);
7 z0 N8 J6 B9 u8 p$ p" e( A frame.setVisible(true);
# p4 i% c5 B% c }$ P/ r1 d7 q. b5 ^7 l" e4 h
} 复制代码 0 H' ~: U8 s! H7 ?: F$ `: n
; |" s: o6 q5 j. }$ B7 y - [3 J# Z* O2 [: P4 M+ t
1 r1 ?6 o2 `$ I }
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