实验报告年级班号学号姓名实验名称：栈的实现及其应用：算术表达式的计算实验日期2016年12月2日计算机科学与技术系2016年制一、实验环境32位操作系统下的Window平台Microsoft Visual C++二、实验目的掌握栈的实现及使用三、实验内容1.实现栈的存储结构2.实现栈的基本操作的有关算法3.利用栈解决*算术表达式求值问题四、数据结构与算法思想描述顺序读取中缀表达式：1、当遇到数字时，将数字入数字栈2、当遇到操作符时，与操作符栈栈顶比较：If(当前操作符优先级大于操作符栈栈顶的优先级)If(非”)”操作符)将当前操作符进操作符栈;ElseWhile(操作符栈栈顶不等于”(“)取操作符栈栈顶及数字栈的两个数进行运算，并将结果压入数字栈;ElseIf(非(“操作符)While(操作符栈栈顶不等于”(“)取操作符栈栈顶及数字栈的两个数进行运算，并将结果压入数字栈;Continue;(直到当前操作符比栈顶操作符优先级大)Else将当前操作符进操作符栈;3、While(操作符栈非空)操作符栈栈顶及数字栈的两个数进行运算，并将结果压入数字栈;4、在数字栈取最后结果并输出。

五、程序清单//10*8^2+16.3+5*(5.2*5+3.01)/4-(-10)+0.1000060+4.00416-40 = 666.666666 //100+(-100)-(-10^2) = 100//(((2016-2017+(((2015-2014)))))) = 0//-1+(((((((((1^0))))))))+100%10^2 = 0#include<iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<iomanip>#include<map>using namespace std;const int MAX = 105;typedef double Type;typedef struct{Type TypeStack[MAX];char charStack[MAX];int TypeTop, charTop;}Stack;//初始化栈void InitStack(Stack *S){S->charTop = S->TypeTop = 0; }//判断charStack是否为空bool IsEmpty_Char(Stack S){return S.charTop == 0;}//判断TypeStack是否为空bool IsEmpty_Type(Stack S){return S.TypeTop == 0;}//判断charStack是否为满bool IsFull_Char(Stack S){return S.charTop == MAX;}//判断TypeStack是否为满bool IsFull_Type(Stack S){return S.TypeTop == MAX;}void Push_Char(Stack *S, char ch){//charStack不为满则入栈，否则输出提示if(!IsFull_Char(*S))S->charStack[S->charTop++] = ch;elsecout << " The CharStack Is Full! " << endl; }void Push_Type(Stack *S, Type a){//TypeStack不为满则入栈，否则输出提示if(!IsFull_Type(*S))S->TypeStack[S->TypeTop++] = a;elsecout << " The TypeStack Is Full! " << endl; }char Pop_Char(Stack *S)if(!IsEmpty_Char(*S)){S->charTop--;return S->charStack[S->charTop];}elsecout << " The CharStack Is Empty! " << endl;return -1;}Type Pop_Type(Stack *S){if(!IsEmpty_Type(*S)){S->TypeTop--;return S->TypeStack[S->TypeTop];}elsecout << " The TypeStack Is Empty! " << endl;return -1;}char Top_Char(Stack S){if(!IsEmpty_Char(S))return S.charStack[--S.charTop];elsecout << " The CharStack Is Empty! " << endl;return -1;}Type Top_Type(Stack S){if(!IsEmpty_Type(S))return S.TypeStack[--S.TypeTop];elsecout << " The TypeStack Is Empty! " << endl;return -1;}Type Calculate(Type left, Type right, char op){Type value = 0;switch(op){case '+': value = left + right; break;case '-': value = left - right; break;case '*': value = left * right; break;case '/': if(right != 0)value = left / right;elsecout << "被除数不能为零!" << endl;break;case '%': i f(right != 0)value = (int)left % (int)right;elsecout << "被余数不能为零!" << endl;break;case '^': value = pow(left,right);/*value = 1;if(right >= 0)while(right--)value *= left;else{right = -right;while(right--)value /= left;}*/}return value;}void Computer(char *mid_equotion, Type len){Type right, left , result;char *p_mid_equotion = mid_equotion;char after_equotion = ' ';map<char,Type> Oper;Oper['#'] = 1; Oper['('] = 2; O per['+'] = 3;Oper['-'] = 3; O per['*'] = 4; Oper['/'] = 4;Oper['%'] = 4; Oper['^'] = 5; Oper[')'] = 6;Stack MyStack;InitStack(&MyStack);Push_Char(&MyStack,'#');char top_oper, current_oper;for(;*p_mid_equotion != '\0';){top_oper = Top_Char(MyStack);current_oper = *p_mid_equotion;if(!Oper[current_oper]){Push_Type(&MyStack,strtod(p_mid_equotion, &p_mid_equotion));continue;}//end ifelse //为操作符{if(Oper[current_oper] > Oper[top_oper]){if(current_oper != ')')Push_Char(&MyStack,current_oper);else{while(top_oper != '('){right = Pop_Type(&MyStack);if(!IsEmpty_Type(MyStack))left = Pop_Type(&MyStack);elseleft = 0;Push_Type(&MyStack,Calculate(left, right, Top_Char(MyStack)));Pop_Char(&MyStack);top_oper = Top_Char(MyStack);}Pop_Char(&MyStack);}//end else}//end ifelse{if(current_oper == '('){Push_Char(&MyStack,current_oper);if(*(p_mid_equotion + 1) == '-')Push_Type(&MyStack,0);}else{right = Pop_Type(&MyStack);if(!IsEmpty_Type(MyStack))left = Pop_Type(&MyStack);elseleft = 0;Push_Type(&MyStack,Calculate(left, right, top_oper));Pop_Char(&MyStack);continue;}}//end else}//end elsep_mid_equotion++;}//end fortop_oper = Pop_Char(&MyStack);while(top_oper != '#'){right = Pop_Type(&MyStack);if(!IsEmpty_Type(MyStack))left = Pop_Type(&MyStack);elseleft = 0;Push_Type(&MyStack,Calculate(left, right, top_oper));top_oper = Pop_Char(&MyStack);}// cout << setprecision(6) << "\nThe Result = " << (result = Pop_Type(&MyStack)) << endl;printf("The Result = %lf\n\n",(result = Pop_Type(&MyStack)));}int main(){char s[MAX] = "";Type i = 0;cout << "请输入你要求值的表达式!(以-1结束)\n";while(cin >> s && strcmp(s,"-1") != 0){Computer(s,strlen(s));cout << "请输入你要求值的表达式!(以-1结束)\n";}return 0;}六、程序执行结果及其分析对“+” , “-” , “*” , “/” , “%” , “^”运算的实现可运算多位数和小数，求余，求平方，括号里包含负数如（-1），及首个数字为负数如-1+1。