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结对项目2.0版
更新内容：
1.在之前基础上实现四个数的四则运算。
2.实现了带有括号的运算。
存在问题：
运算过程中偶尔会有-nan(ind)的结果，还在找原因。
部分代码如下：
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(); 226 }
这部分为计算器，主体来自http://blog.csdn.net/bhq2010/article/details/7516369，做了些必要的修改。
1 int N = 0, i = 0; 2 int optr1 = 0, optr2 = 0, optr3 = 0, kuohao = 0; 3 int num1, num2, num3, num4; 4 5 double answer1, answer2; 6 struct biaodashi string[15]; 7 char str[] = "+-*/", ch; 8 9 srand((unsigned)time(0));10 11 printf("请输入本次测试的题目数量："); //手动输入题目数量12 scanf("%d", &N);13 ch = getchar();14 15 printf("保留两位小数\n"); //输入结果如果有小数，保留两位小数
部分主函数代码。
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 }
这部分为不带括号的出题方式，比例大小为60%
else { printf("题号：%d\n", i + 1); switch (kuohao) { case 1: printf("(%d %c %d) %c %d %c %d=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = ‘(‘; string[1].word = num1; string[2].word = str[optr1]; string[3].word = num2; string[4].word = ‘)‘; string[5].word = str[optr2]; string[6].word= num3; string[7].word = str[optr3]; string[8].word = num4; string[9].word = ‘=‘; break; case 2: printf("(%d %c %d %c %d) %c %d=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = ‘(‘; string[1].word = num1; string[2].word = str[optr1]; string[3].word = num2; string[4].word = str[optr2]; string[5].word = num3; string[6].word = ‘)‘; string[7].word = str[optr3]; string[8].word = num4; string[9].word = ‘=‘; break; case 3: printf("%d %c (%d %c %d) %c %d=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = num1; string[1].word = str[optr1]; string[2].word = ‘(‘; string[3].word = num2; string[4].word = str[optr2]; string[5].word = num3; string[6].word = ‘)‘; string[7].word = str[optr3]; string[8].word = num4; string[9].word = ‘=‘; break; case 4: printf("%d %c (%d %c %d %c %d)=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = num1; string[1].word = str[optr1]; string[2].word = ‘(‘; string[3].word = num2; string[4].word = str[optr2]; string[5].word = num3; string[6].word = str[optr3]; string[7].word = num4; string[8].word = ‘)‘; string[9].word = ‘=‘; break; case 5: printf("%d %c %d %c (%d %c %d)=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = num1; string[1].word = str[optr1]; string[2].word = num2; string[3].word= str[optr2]; string[4].word = ‘(‘; string[5].word = num3; string[6].word = str[optr3]; string[7].word = num4; string[8].word = ‘)‘; string[9].word = ‘=‘; break; case 6: printf("(%d %c %d) %c (%d %c %d)=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = ‘(‘; string[1].word = num1; string[2].word = str[optr1]; string[3].word = num2; string[4].word = ‘)‘; string[5].word = str[optr2]; string[6].word = ‘(‘; string[7].word = num3; string[8].word = str[optr3]; string[9].word = num4; string[10].word = ‘)‘; string[11].word = ‘=‘; break; case 7: printf("(%d %c (%d %c %d)) %c %d=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = ‘(‘; string[1].word = num1; string[2].word = str[optr1]; string[3].word = ‘(‘; string[4].word = num2; string[5].word = str[optr2]; string[6].word = num3; string[7].word = ‘)‘; string[8].word = ‘)‘; string[9].word = str[optr3]; string[10].word = num4; string[11].word = ‘=‘; break; case 8: printf("%d %c (%d %c (%d %c %d))=?\n", num1, str[optr1], num2, str[optr2], num3, str[optr3], num4); string[0].word = num1; string[1].word = str[optr1]; string[2].word = ‘(‘; string[3].word = num2; string[4].word = str[optr2]; string[5].word= ‘(‘; string[6].word = num3; string[7].word = str[optr3]; string[8].word = num4; string[9].word = ‘)‘; string[10].word = ‘)‘; string[11].word = ‘=‘; break; default: printf("错误\n"); } answer1 = Calculate(string); scanf("%lf", &answer2); ch = getchar(); if ((fabs(answer2-answer1))<0.01) printf("正确\n"); else printf("错误 正确答案是：%-10.2lf\n", answer1); }
带括号的出题方式采用的枚举法，所以略长，比例为40%
输出结果：
结果：可以看到第6题出现了-nan(ind)的问题，百度一下意思是not a number，应该是计算过程中出了问题，但是还没找到。
结对项目2.0版
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