24点游戏 程序(二)

前面既然确定了后缀表达式的序列。

那就可以开始遍历了。

#include <iostream>#include <stack>#include <algorithm>#include <string>#include <cmath>#include <time.h>using namespace std;struct TOperData{    int operType;    double data;};//检查计算的参数是否合法，也可以用于过滤一些重复的表达式bool checkjisuan(double a, double b, int op){    //处理除以0的情况    if (op == '/' && b < 1e-6 && b > -1e-6)        return false;    return true;}//求值double jisuan(double a, double b, int op){    switch(op)    {    case '+':        return a + b;    case '-':        return a - b;    case '*':        return a * b;    case '/':        return a / b;    default:        return -1;    }}//计算表达式的值double process(TOperData data[]){    int len = 7;    stack<TOperData> suffix;    for (int i = 0; i < len; i++)    {        if (data[i].operType == 0)        {            suffix.push(data[i]);        }        else if (data[i].operType > 0)        {            if (suffix.empty())                return false;            if (suffix.top().operType == 0)            {                double a = suffix.top().data;                suffix.pop();                if (!suffix.empty() && suffix.top().operType == 0)                {                    double b = suffix.top().data;                    suffix.pop();                    if (!checkjisuan(b,a,data[i].operType)) return -1;                    double c = jisuan(b,a,data[i].operType);                    TOperData opdata;                    opdata.operType = 0;                    opdata.data = c;                    suffix.push(opdata);                }                else                {                    return -1;                }            }            else            {                return -1;            }        }    }    if (suffix.empty()) return -1;    if (suffix.top().operType == 0)    {        double r = suffix.top().data;        suffix.pop();        if (suffix.empty()) return r;    }    return -1;}//打印成功的结果void printResult(TOperData computeData[]){    for (int i = 0; i < 7; i++)    {        if (computeData[i].operType == 0)        {            cout<<computeData[i].data<<" ";        }        else        {            cout<<(char)computeData[i].operType<<" ";        }    }    cout<<endl;}int start(double data[]){    int shunxu[][7] = {1,1,1,1,2,2,2 ,                        1,1,1,2,1,2,2 ,                        1,1,1,2,2,1,2 ,                        1,1,2,1,1,2,2 ,                        1,1,2,1,2,1,2                      };    int operAll[] = {'+','-','*','/'};    TOperData computeData[7];    double copydata[4];    copydata[0] = data[0];    copydata[1] = data[1];    copydata[2] = data[2];    copydata[3] = data[3];    //5个后缀表达式遍历    for (int m = 0; m < 5; m++)    {        do        {            int oper[3];            for(int n = 0; n < 4; n++)            {                oper[0] = operAll[n];                for(int nn = 0; nn < 4; nn++)                {                    oper[1] = operAll[nn];                    for(int nnn = 0; nnn < 4; nnn++)                    {                        oper[2] = operAll[nnn];                        int j = 0;                        int k = 0;                        for (int i = 0; i < 7; i++)                        {                            if (shunxu[m][i] == 1)                            {                                computeData[i].operType = 0;                                computeData[i].data = data[j++];                            }                            else if (shunxu[m][i] == 2)                            {                                computeData[i].operType = oper[k++];                            }                        }                        double r = process(computeData);                        if (r-24 > -1e-6 && r-24 < 1e-6)                        {                            cout<<copydata[0]<<" "<<copydata[1]<<" "<<copydata[2]<<" "<<copydata[3]<<",";                            printResult(computeData);                            //return 1;如果只要求一个结果，这个地方就可以返回了                        }                    }                }            }        }        while ( next_permutation (data,data+4) );    }    return 0;}int main(void){    long beginTime =clock();//获得开始时间，单位为毫秒    double data[4];    for (int i = 1; i< 14; i++)    {        data[0] = i;        for (int ii = i; ii< 14; ii++)        {            data[1] = ii;            for (int iii = ii; iii< 14; iii++)            {                data[2] = iii;                for (int iiii = iii; iiii< 14; iiii++)                {                    data[3] = iiii;                    double copydata[4];                    copydata[0] = data[0];                    copydata[1] = data[1];                    copydata[2] = data[2];                    copydata[3] = data[3];                    start(copydata);                }            }        }    }    long endTime=clock();//获得结束时间    cout<<"time:"<<endTime-beginTime<<endl;    return 0;}

这样执行得到50000个结果，耗时30s左右

仔细看这些结果，可以发现有一些重复的。

接下来要简单消除一下重复，以及为了结果更直观，把后缀表达式改成中缀表达式。