面：24点游戏

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输入4个数，求解得到24点，并输出相应的表达式。 使用面向对象开发，只能用整形，比如输入6 6 6 6 ，则输出6+6+6+6=24

测试的时候我会用3 3 8 8，也可能会有5 5 5 1 来测试

3 3 8 8
(((3+8)/3)*8) = 24

3,3,8,8算24点的算法有：

1，8/(3-8/3)=24

2，8/(3-(8/3))=24

8÷(3-8÷3)=8÷1/3=24

加减乘除24点（难度系数：2颗星）

计算24点有什么窍门或技巧吗?

用穷举法列出24点的解

24点游戏算法（C语言实现）

“速算24″算法思路

关于24点游戏的编程思路与基本算法

面：24点游戏48行计算24点C语言代码

面：24点游戏

4X6=24

3X8=24

a  _   b  _   c   _   d:

C41 * C41 *C41 * A44这就是错的原因，没有考虑全部的排列方式？

最新版本，

#include <iostream> #include <stdio.h> #include <string.h> using namespace std; #define OP(o) do { a[1].n = a[0].n o a[1].n; \ sprintf(a[1].s, "(%s"#o"%s)", a[0].s, strcpy(t.s, a[1].s)); \ if (dfs(n - 1, a + 1, r)) return 1; \ a[1] = temp; } while (0) typedef struct data { double n;//存中间结果 char s[8 * 7];//存表达式 } data; typedef int Ptype;//para type data num[4], t; class Solution{ public: Solution(); ~Solution(); int dfs(int n, data a[], int r);//TODO }; Solution::Solution(){ //sth } Solution::~Solution(){ //sth } int Solution::dfs( int n, data a[], int r ) {// n,数字个数； a[] 存输入的n个数，r 要得到的结果 int i, j; if ( n == 1 ) if ( a[0].n-r > 1e-8 || a[0].n-r < -1e-8 ) return 0;//　!=0 else return 1; for ( i = 0; i < n; ++i ) { swap(a[i], a[0] ); for ( j = 1; j < n; ++j ) { swap(a[j], a[1] );//指定为data类型 data temp = a[1]; OP( + ); OP( - ); OP( * ); if ( a[1].n > 1e-8 || a[1].n < -1e-8 ) OP( / ); swap(a[j], a[1] ); } swap(a[i], a[0] ); } return 0; } int main() { int r = 24, i; printf("输入4个数字，以空格隔开：\n"); for ( i = 0; i < 4; ++i ) { scanf( "%s", num[i].s ); sscanf( num[i].s, "%lf", &num[i].n ); } Solution* ans=new Solution; if (ans->dfs( 4, num, r ) ) printf( "%s = %d\n", num[3].s, r ); else printf( "不存在相应的24点的组合\n" ); return 0; }

混合：可正常测试出3 3 8 8和 5 5 5 1了。

又被说，，，

#include <iostream> #include <stdio.h> #include <string.h> using namespace std; #define SWAP(tp, a, b) do {tp v=a; a=b; b=v;} while (0) #define OP(o) do { a[1].n = a[0].n o a[1].n; \ sprintf(a[1].s, "(%s"#o"%s)", a[0].s, strcpy(t.s, a[1].s)); \ if (dfs(n - 1, a + 1, r)) return 1; \ a[1] = temp; } while (0) typedef struct data { double n;//用int存的话会得不出正常答案 char s[8 * 7];//存表达式 } data; typedef int Ptype;//para type data num[4], t; class Solution{ public: Solution(); ~Solution(); int dfs(int n, data a[], int r);//TODO int TotalCount(Ptype a, Ptype b, Ptype c, Ptype d);//C4 1 private: void Calculate(int fl,Ptype n,Ptype &sum, char &s);//A3 3 int Is24(Ptype a, Ptype b, Ptype c, Ptype d); int Count(Ptype a, Ptype b, Ptype c, Ptype d); protected: //data num[4], t; }; Solution::Solution(){ //sth } Solution::~Solution(){ //sth } void Solution::Calculate(int fl,Ptype n,Ptype &sum, char &s) { //+-*/ 1234 switch (fl) {//flag case 1: sum += n; s='+'; break; case 2: sum -= n; s='-'; break; case 3: sum *= n; s='*'; break; case 4: { if (sum%n) sum = 9999;//指向一个得不到24的值 sum /= n;//else } break; default: break; } } int Solution::Is24(Ptype a, Ptype b, Ptype c, Ptype d) { //a b c d 4个数中间分别都有+ - * / 四种运算符的可能性 暴力枚举 Ptype sum; char s1,s2,s3;// 存运算符 for (int j = 1; j <= 4; j++) {//第1个位置的符号 for (int k = 1; k <= 4; k++) {//第2个位置 for (int l = 1; l <= 4; l++) {//第3个位置 sum = 0; sum += a; Calculate(j,b,sum,s1); Calculate(k,c,sum,s2); Calculate(l,d,sum,s3); if (sum == 24) {//格式化输出 printf("组合：((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); return 1; } } } } return 0; } int Solution::Count(Ptype a, Ptype b, Ptype c, Ptype d) {//A3 3 return Is24(a, b, c, d) + Is24(a, b, d, c) + Is24(a, c, b, d) + Is24(a, c, d, b) + Is24(a, d, b, c) + Is24(a, d, c, b); } int Solution::TotalCount(Ptype a, Ptype b, Ptype c, Ptype d) {//C4 1 return Count(a, b, c, d) + Count(b, a, c, d) + Count(c, a, b, d) + Count(d, a, b, c); } int Solution::dfs( int n, data a[], int r ) {// n,数字个数； a[] 存输入的n个数，r 要得到的结果 int i, j; if ( n == 1 ) if ( a[0].n-r > 1e-8 || a[0].n-r < -1e-8 ) return 0; else return 1; for ( i = 0; i < n; ++i ) { SWAP( data, a[i], a[0] ); for ( j = 1; j < n; ++j ) { SWAP( data, a[j], a[1] );//指定为data类型 data temp = a[1]; OP( + ); OP( - ); OP( * ); if ( a[1].n > 1e-8 || a[1].n < -1e-8 ) OP( / ); SWAP( data, a[j], a[1] ); } SWAP( data, a[i], a[0] ); } return 0; } int main() { /* * int a, b, c, d; * cout<<"请输入4个数，以空格隔开:"<<endl; * cin >> a >> b >> c >> d; * * Solution* ans=new Solution; * int ct = ans->TotalCount(a, b, c, d);//A4 4 * delete ans; * * if (ct == 0) cout << "不存在相应的24点的组合" << endl; * cout<<"总数："<<ct<<endl; * return 0; */ int r = 24, i; printf("输入4个数字，以空格隔开：\n"); for ( i = 0; i < 4; ++i ) { scanf( "%s", num[i].s ); sscanf( num[i].s, "%lf", &num[i].n ); } Solution* ans=new Solution; if (ans->dfs( 4, num, r ) ) printf( "%s = %d\n", num[3].s, r ); else printf( "不存在相应的24点的组合\n" ); return 0; }

被大佬嫌弃，，，， 可哥，不想改了

#include <iostream> #include <stdio.h> typedef int Ptype;//para type using namespace std; class Solution{ public: Solution(); ~Solution(); int TotalCount(Ptype a, Ptype b, Ptype c, Ptype d);//C4 1 private: void Calculate(int i,Ptype n,Ptype &sum, char &s);//A3 3 int Is24(Ptype a, Ptype b, Ptype c, Ptype d); int Count(Ptype a, Ptype b, Ptype c, Ptype d); //int Is24_DFS();//TODO }; Solution::Solution(){ //sth } Solution::~Solution(){ //sth } void Solution::Calculate(int i,Ptype n,Ptype &sum, char &s) { //+-*/ 1234 switch (i) { case 1: sum += n; s='+'; break; case 2: sum -= n; s='-'; break; case 3: sum *= n; s='*'; break; case 4: { if (sum%n) sum = 9999;//指向一个得不到24的值 sum /= n;//else } break; default: break; } } int Solution::Is24(Ptype a, Ptype b, Ptype c, Ptype d) { //a b c d 4个数中间分别都有+ - * / 四种运算符的可能性 暴力枚举 Ptype sum; char s1,s2,s3;// 存运算符 for (int j = 1; j <= 4; j++) {//第1个位置的符号 for (int k = 1; k <= 4; k++) {//第2个位置 for (int l = 1; l <= 4; l++) {//第3个位置 sum = 0; sum += a; Calculate(j,b,sum,s1); Calculate(k,c,sum,s2); Calculate(l,d,sum,s3); if (sum == 24) {//格式化输出 printf("组合：((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); return 1; } } } } return 0; } int Solution::Count(Ptype a, Ptype b, Ptype c, Ptype d) {//A3 3 return Is24(a, b, c, d) + Is24(a, b, d, c) + Is24(a, c, b, d) + Is24(a, c, d, b) + Is24(a, d, b, c) + Is24(a, d, c, b); } int Solution::TotalCount(Ptype a, Ptype b, Ptype c, Ptype d) {//C4 1 return Count(a, b, c, d) + Count(b, a, c, d) + Count(c, a, b, d) + Count(d, a, b, c); } int main() { int a, b, c, d; cout<<"请输入4个数，以空格隔开:"<<endl; cin >> a >> b >> c >> d; Solution* ans=new Solution; int ct = ans->TotalCount(a, b, c, d);//A4 4 delete ans; if (ct == 0) cout << "不存在相应的24点的组合" << endl; cout<<"总数："<<ct<<endl; return 0; }

#include <iostream> #include <stdio.h> #include <math.h> using namespace std; class Solution{ public: Solution(); ~Solution(); int TotalCount(int a, int b, int c, int d);//get the total num private: void AddUp(int i,int n,int &sum, char &s);//add nums up int Is24(int a, int b, int c, int d); int Count(int a, int b, int c, int d); //typedef XXX if needed };//抽到头文件,大的时候 Solution::Solution(){ //sth } Solution::~Solution(){ //sth } void Solution::AddUp(int i,int n,int &sum, char &s) {//计算 //+-*/ 1234 switch (i) { case 1: sum += n; s='+'; break; case 2: sum -= n; s='-'; break; case 3: sum *= n; s='*'; break; case 4: { if (sum%n) { sum = 9999;//指向一个得不到24的值 } else { sum /= n; } }break; default: break; } } int Solution::Is24(int a, int b, int c, int d) { int sum; char s1,s2,s3;// 存运算符 //a b c d 4个数中间分别都有+ - * / 四种运算符的可能性 暴力枚举 for (int j = 1; j <= 4; j++) {//第1个位置的符号 for (int k = 1; k <= 4; k++) {//第2个位置的符号 for (int l = 1; l <= 4; l++) {//第3个位置的符号 sum = 0; sum += a; AddUp(j,b,sum,s1); AddUp(k,c,sum,s2); AddUp(l,d,sum,s3); if (sum == 24) {//判断并格式化输出 printf("组合：((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); return 1; } } } } return 0; } int Solution::Count(int a, int b, int c, int d) { return Is24(a, b, c, d) + Is24(a, b, d, c) + Is24(a, c, b, d) + Is24(a, c, d, b) + Is24(a, d, b, c) + Is24(a, d, c, b);//各种排列 } int Solution::TotalCount(int a, int b, int c, int d) { return Count(a, b, c, d) + Count(b, a, c, d) + Count(c, a, b, d) + Count(d, a, b, c);//排列 } int main() { int a, b, c, d; int count = 0; cout<<"请输入4个数，以空格隔开:"<<endl; cin >> a >> b >> c >> d; Solution* ans=new Solution; count = ans->TotalCount(a, b, c, d); if (count == 0) cout << "不存在相应的24点的组合" << endl; cout<<"总数："<<count<<endl; return 0; }

修改历史：

我是垃圾，，，写不出来啊，，，抄：

面：24点游戏

#include <iostream> //#include <cstdio> //#include <cmath> #include <stdio.h> #include <math.h> using namespace std; //+-*/ 1234 unsigned long long int xx[1001]; int x=1; int pd; int f1(int a, int b, int c, int d) { int sum; pd=0; char s1,s2,s3; for (int j = 1; j <= 4; j++) { for (int k = 1; k <= 4; k++) { for (int l = 1; l <= 4; l++) { sum = 0; sum += a; switch (j) { case 1:sum += b; break; case 2:sum -= b; break; case 3:sum *= b; break; case 4: { if (sum%b) { sum = 9999; } else { sum /= b; } }break; default: break; } switch (k) { case 1:sum += c; break; case 2:sum -= c; break; case 3:sum *= c; break; case 4: { if (sum%c) { sum = 9999; } else { sum /= c; } }break; default: break; } switch (l) { case 1:sum += d; break; case 2:sum -= d; break; case 3:sum *= d; break; case 4: { if (sum%d) { sum = 9999; } else { sum /= d; } }break; default: break; } switch (j) { case 1:s1='+';break; case 2:s1='-';break; case 3:s1='*';break; case 4:s1='/';break; } switch (k) { case 1:s2='+';break; case 2:s2='-';break; case 3:s2='*';break; case 4:s2='/';break; } switch (l) { case 1:s3='+';break; case 2:s3='-';break; case 3:s3='*';break; case 4:s3='/';break; } if (sum == 24) { if (x==0) { xx[x]=a*+b*+c*10000+d*1000+j*100+k*10+l; x++; printf("((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); } else { for (int i=1;i<=x;i++) { if ((a*+b*+c*10000+d*1000+j*100+k*10+l)==xx[i]) { pd=1; break; } } if (pd==0) { xx[x]=a*+b*+c*10000+d*1000+j*100+k*10+l; x++; printf("((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); } } return 1; } } } } return 0; } int f2(int a, int b, int c, int d) { int sum = 0; sum = f1(a, b, c, d) + f1(a, b, d, c) + f1(a, c, b, d) + f1(a, c, d, b) + f1(a, d, b, c) + f1(a, d, c, b); if (sum != 0) { cout << "存在" << endl; return 1; } else { cout << "不存在" << endl; return 0; } } int main() { int a, b, c, d; int sum; cout<<"请输入4个数，以空格隔开如：3 3 8 9"<<endl; cin >> a >> b >> c >> d; sum = f2(a, b, c, d) + f2(b, a, c, d) + f2(c, a, b, d) + f2(d, a, b, c); if (sum == 0) { cout << "不存在相应的24点的组合" << endl; } return 0; }

自己注释：

#include <iostream> #include <stdio.h> #include <math.h> using namespace std; //+-*/ 1234 unsigned long long int xx[1001];// 保存值的数组 int x=1;// int pd;//判断 int f1(int a, int b, int c, int d) { int sum; pd=0; char s1,s2,s3;// 存运算符 //a b c d 4个数中间分别都有+ - * / 四种运算符的可能性 暴力枚举 for (int j = 1; j <= 4; j++) {//第1个位置的符号 for (int k = 1; k <= 4; k++) {//第2个位置的符号 for (int l = 1; l <= 4; l++) {//第3个位置的符号 sum = 0; sum += a; switch (j) { case 1:sum += b; break; case 2:sum -= b; break; case 3:sum *= b; break; case 4: { if (sum%b) { sum = 9999; } else { sum /= b; } }break; default: break; } switch (k) { case 1:sum += c; break; case 2:sum -= c; break; case 3:sum *= c; break; case 4: { if (sum%c) { sum = 9999; } else { sum /= c; } }break; default: break; } switch (l) { case 1:sum += d; break; case 2:sum -= d; break; case 3:sum *= d; break; case 4: { if (sum%d) {//不能被整除 sum = 9999;//不能被整除，意味着得不到24 } else {//能被整除 sum /= d; } }break; default: break; } //存符号 switch (j) { case 1:s1='+';break; case 2:s1='-';break; case 3:s1='*';break; case 4:s1='/';break; } switch (k) { case 1:s2='+';break; case 2:s2='-';break; case 3:s2='*';break; case 4:s2='/';break; } switch (l) { case 1:s3='+';break; case 2:s3='-';break; case 3:s3='*';break; case 4:s3='/';break; } if (sum == 24) {//判断并格式化输出 if (x==0) {//第1个等于24的时候进去？永远不会执行？ xx[x]=a*+b*+c*10000+d*1000+j*100+k*10+l;//存组合及符号 x++; printf("x==0,,,,((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); } else { for (int i=1;i<=x;i++) { if ((a*+b*+c*10000+d*1000+j*100+k*10+l)==xx[i]) { pd=1; break; } } if (pd==0) { xx[x]=a*+b*+c*10000+d*1000+j*100+k*10+l; x++; printf("pd==0,,,((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); } } return 1; } } } } return 0; } int f2(int a, int b, int c, int d) { int sum = 0; sum = f1(a, b, c, d) + f1(a, b, d, c) + f1(a, c, b, d) + f1(a, c, d, b) + f1(a, d, b, c) + f1(a, d, c, b);//各种排列 if (sum != 0) { //cout << "存在" << endl; return 1; } else { //cout << "不存在" << endl; return 0; } } int main() { int a, b, c, d; int sum; cout<<"请输入4个数，以空格隔开:"<<endl; cin >> a >> b >> c >> d; sum = f2(a, b, c, d) + f2(b, a, c, d) + f2(c, a, b, d) + f2(d, a, b, c); if (sum == 0) { cout << "不存在相应的24点的组合" << endl; } return 0; }

修改：

看不懂里面写的一些东西是干什么的，，，

不是很懂大佬之前的意思，，

删，

#include <iostream> #include <stdio.h> #include <math.h> using namespace std; //+-*/ 1234 void SwitchFunc(int i,int c,int &sum, char &s2) { switch (i) { case 1: sum += c; s2='+'; break; case 2: sum -= c; s2='-'; break; case 3: sum *= c; s2='*'; break; case 4: sum /= c; s2='/'; break; default: break; } } int f1(int a, int b, int c, int d) { int sum; char s1,s2,s3;// 存运算符 //a b c d 4个数中间分别都有+ - * / 四种运算符的可能性 暴力枚举 for (int j = 1; j <= 4; j++) {//第1个位置的符号 for (int k = 1; k <= 4; k++) {//第2个位置的符号 for (int l = 1; l <= 4; l++) {//第3个位置的符号 sum = 0; sum += a; SwitchFunc(j,b,sum,s1); SwitchFunc(k,c,sum,s2); SwitchFunc(l,d,sum,s3); if (sum == 24) {//判断并格式化输出 printf("pd==waimian,,,((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); return 1; } } } } return 0; } int f2(int a, int b, int c, int d) { int sum = 0; sum = f1(a, b, c, d) + f1(a, b, d, c) + f1(a, c, b, d) + f1(a, c, d, b) + f1(a, d, b, c) + f1(a, d, c, b);//各种排列 return sum; } int main() { int a, b, c, d; int sum; cout<<"请输入4个数，以空格隔开:"<<endl; cin >> a >> b >> c >> d; sum = f2(a, b, c, d) + f2(b, a, c, d) + f2(c, a, b, d) + f2(d, a, b, c);//为啥要这样写？？？不懂 if (sum == 0) { cout << "不存在相应的24点的组合" << endl; } return 0; }

改，

#include <iostream> #include <stdio.h> #include <math.h> using namespace std; //+-*/ 1234 void SwitchFunc(int i,int n,int &sum, char &s) { switch (i) { case 1: sum += n; s='+'; break; case 2: sum -= n; s='-'; break; case 3: sum *= n; s='*'; break; case 4: sum /= n; s='/'; break; default: break; } } int Is24(int a, int b, int c, int d) { int sum; char s1,s2,s3;// 存运算符 //a b c d 4个数中间分别都有+ - * / 四种运算符的可能性 暴力枚举 for (int j = 1; j <= 4; j++) {//第1个位置的符号 for (int k = 1; k <= 4; k++) {//第2个位置的符号 for (int l = 1; l <= 4; l++) {//第3个位置的符号 sum = 0; sum += a; SwitchFunc(j,b,sum,s1); SwitchFunc(k,c,sum,s2); SwitchFunc(l,d,sum,s3); if (sum == 24) {//判断并格式化输出 printf("组合：((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); return 1; } } } } return 0; } int Count(int a, int b, int c, int d) { return Is24(a, b, c, d) + Is24(a, b, d, c) + Is24(a, c, b, d) + Is24(a, c, d, b) + Is24(a, d, b, c) + Is24(a, d, c, b);//各种排列 } int main() { int a, b, c, d; int count = 0; cout<<"请输入4个数，以空格隔开:"<<endl; cin >> a >> b >> c >> d; count = Count(a, b, c, d) + Count(b, a, c, d) + Count(c, a, b, d) + Count(d, a, b, c); if (count == 0) cout << "不存在相应的24点的组合" << endl; cout<<"总数："<<count<<endl; return 0; }

对象，

#include <iostream> #include <stdio.h> #include <math.h> using namespace std; class Solution{ public: void AddUp(int i,int n,int &sum, char &s) {//计算 //+-*/ 1234 switch (i) { case 1: sum += n; s='+'; break; case 2: sum -= n; s='-'; break; case 3: sum *= n; s='*'; break; case 4: sum /= n; s='/'; break; default: break; } } int Is24(int a, int b, int c, int d) { int sum; char s1,s2,s3;// 存运算符 //a b c d 4个数中间分别都有+ - * / 四种运算符的可能性 暴力枚举 for (int j = 1; j <= 4; j++) {//第1个位置的符号 for (int k = 1; k <= 4; k++) {//第2个位置的符号 for (int l = 1; l <= 4; l++) {//第3个位置的符号 sum = 0; sum += a; AddUp(j,b,sum,s1); AddUp(k,c,sum,s2); AddUp(l,d,sum,s3); if (sum == 24) {//判断并格式化输出 printf("组合：((%d%c%d)%c%d)%c%d==24\n",a,s1,b,s2,c,s3,d); return 1; } } } } return 0; } int Count(int a, int b, int c, int d) { return Is24(a, b, c, d) + Is24(a, b, d, c) + Is24(a, c, b, d) + Is24(a, c, d, b) + Is24(a, d, b, c) + Is24(a, d, c, b);//各种排列 } };//class Solution int main() { int a, b, c, d; int count = 0; cout<<"请输入4个数，以空格隔开:"<<endl; cin >> a >> b >> c >> d; Solution* ans=new Solution; count = ans->Count(a, b, c, d) +ans-> Count(b, a, c, d) + ans->Count(c, a, b, d) +ans-> Count(d, a, b, c);//排列 if (count == 0) cout << "不存在相应的24点的组合" << endl; cout<<"总数："<<count<<endl; return 0; }