1 //运算符栈的长度    2 #define OPSTACK_LENGTH 5    3 //操作数栈的长度    4 #define NUMSTACK_LENGTH 100    5 //输入串的最大长度    6 #define MAX_STRING_LENGTH 100   7 #pragma warning(disable:4996)  8 //表达式结构体  9 struct biaodashi 10 { 11     char word; 12 }; 13 //运算符结构体   14 struct operatorStruct 15 { 16     //运算符名称   17     char name; 18     //优先级   19     int priority; 20     //目数，即操作数个数，例如单目运算符为1，双目运算符2   21     int opnum; 22 }; 23  24 typedef struct operatorStruct OPERATOR; 25  26 //运算符栈   27 OPERATOR opStack[OPSTACK_LENGTH]; 28 //运算符栈顶指针   29  30 //操作数栈   31 double numStack[NUMSTACK_LENGTH]; 32 //操作数栈顶指针   33 int opStackTop = -1; 34 int numStackTop = -1; 35  36 //获取一个字符所代表的运算符的优先级   37 int getPriority(char name) 38 { 39     if (name == ‘(‘ || name == ‘)‘) 40     { 41         return 0; 42     } 43     if (name == ‘!‘) 44     { 45         return 3; 46     } 47     if (name == ‘*‘ || name == ‘/‘) 48     { 49         return 2; 50     } 51     if (name == ‘+‘ || name == ‘-‘) 52     { 53         return 1; 54     } 55      56 } 57 //获取一个字符所代表的运算符的目数   58 int getOpNum(char name) 59 { 60     if (name == ‘*‘ || name == ‘/‘ || name == ‘+‘ || name == ‘-‘) 61     { 62         return 2; 63     } 64     if (name == ‘!‘) 65     { 66         return 1; 67     } 68     if (name == ‘(‘ || name == ‘)‘) 69     { 70         return 0; 71     } 72  73 } 74  75 //运算符压栈   76 void pushOperator(OPERATOR op) 77 { 78     if (opStackTop < OPSTACK_LENGTH - 1) 79     { 80         opStack[++opStackTop] = op; 81     } 82 } 83 //运算符出栈   84 OPERATOR popOperator() 85 { 86     if (opStackTop >= 0) 87     { 88         return opStack[opStackTop--]; 89     } 90 } 91 //操作数压栈   92 void pushNumber(double num) 93 { 94     if (numStackTop < NUMSTACK_LENGTH - 1) 95     { 96         numStack[++numStackTop] = num; 97     } 98 } 99 //操作数出栈  100 double popNumber()101 {102     if (numStackTop >= 0)103     {104         return numStack[numStackTop--];105     }106 }107 108 //从操作数栈中弹出两个操作数，完成一次双目运算  109 double opertate2Num(OPERATOR op)110 {111     double num2 = popNumber();112     double num1 = popNumber();113     if (op.name == ‘+‘)114     {115         return num1 + num2;116     }117     if (op.name == ‘-‘)118     {119         return num1 - num2;120     }121     if (op.name == ‘*‘)122     {123         return num1 * num2;124     }125     if (op.name == ‘/‘)126     {127         return num1 / num2;128     }129 }130 //从操作数栈中弹出一个操作数，完成一次单目运算  131 double opertate1Num(OPERATOR op)132 {133     double num = popNumber();134     if (op.name == ‘!‘)135     {136         double result = 1;137         while (num > 1)138         {139             result *= num;140             num--;141         }142         return result;143     }144 145 }146 //完成一次运算  147 double operate(OPERATOR op)148 {149     if (op.opnum == 1)150     {151         return opertate1Num(op);152     }153     else if (op.opnum == 2)154     {155         return opertate2Num(op);156     }157 158 }159 //四则运算计算器160 double Calculate(struct biaodashi *string)161 {162     163     int i;164     OPERATOR op, topOp;//op为从当前输入串中提取的一个运算符，topOp为运算符栈栈顶的运算符  165 166     topOp.name = ‘#‘;167     topOp.priority = 0;168     topOp.opnum = 0;169     pushOperator(topOp);//压入#作为初始运算符  170 171     for (i = 0; string[i].word != ‘=‘;i++)172     {173         //从输入串中取出一个字符作为开始，进行处理，直到表达式结束  174         if (string[i].word!=‘+‘ && string[i].word != ‘-‘&& string[i].word != ‘*‘&& string[i].word != ‘/‘&& string[i].word != ‘(‘&& string[i].word != ‘)‘)175         {176             //如果是操作数，将整个操作数提取出来，压入操作数栈 177             pushNumber((double)(string[i].word));178         }179         else180         {181             op.name = string[i].word;182             op.priority = getPriority(string[i].word);183             op.opnum = getOpNum(string[i].word);184             topOp = popOperator();185             if (op.name == ‘(‘)186             {187                 //如果是‘(‘，将从栈顶弹出的运算符压回栈内，并将当前运算符则压栈  188                 pushOperator(topOp);189                 pushOperator(op);190             }191             else if (op.name == ‘)‘)192             {193                 //如果是‘)‘，则进行运算，每次运算结果作为一个操作数压入操作数栈，直到将‘(‘弹出运算符栈  194                 while (topOp.name != ‘(‘)195                 {196                     pushNumber(operate(topOp));197                     topOp = popOperator();198                 }199             }200             else201             {202                 //如果是普通运算符  203                 if (topOp.name != ‘#‘ && op.priority <= topOp.priority)204                 {205                     //如果运算符栈非空，且当前运算符的优先级大于栈顶运算符，则进行一次运算，将结果压入操作数栈  206                     pushNumber(operate(topOp));207                 }208                 else209                 {210                     //否则将从栈顶弹出的运算符压回  211                     pushOperator(topOp);212                 }213                 //将当前运算符压栈  214                 pushOperator(op);215             }216         }217 218     }219     //完成栈内剩余的运算  220     while ((topOp = popOperator()).name != ‘#‘)221     {222         pushNumber(operate(topOp));223     }224     //操作数栈中剩下的最后一个数即为结果  225     return popNumber();            }

 1     for (i = 0; i < N; i++) 2     { 3         num1 = rand() % 100 + 1;                       //生成随机数 4         num2 = rand() % 100 + 1; 5         num3 = rand() % 100 + 1; 6         num4 = rand() % 100 + 1; 7         optr1 = rand() % 4; 8         optr2 = rand() % 4; 9         optr3 = rand() % 4;10         kuohao = rand() % 8;11         if (num1 > 40)12         {13             printf("题号：%d\n", i + 1);14             printf("%d %c %d %c %d %c %d=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4);15             string[0].word = num1;16             string[1].word = str[optr1];17             string[2].word = num2;18             string[3].word = str[optr2];19             string[4].word = num3;20             string[5].word = str[optr3];21             string[6].word = num4;22             string[7].word = ‘=‘;23             answer1 = Calculate(string);24             scanf("%lf", &answer2);25             ch = getchar();26             if ((fabs(answer2-answer1))<0.01)27                 printf("正确\n");28             else printf("错误 正确答案是：%-10.2lf\n", answer1);29         }