長沙理工大學(xué)數(shù)據(jù)結(jié)構(gòu)棧的實(shí)現(xiàn)及應(yīng)用算術(shù)表達(dá)式求值實(shí)驗(yàn)報(bào)告

1、批閱實(shí) 驗(yàn) 報(bào) 告 年級 班號 學(xué)號 姓名 實(shí)驗(yàn)名稱： 棧的實(shí)現(xiàn)及其應(yīng)用：算術(shù)表達(dá)式的計(jì)算 實(shí)驗(yàn)日期 2016年12月 2日實(shí)驗(yàn)報(bào)告撰寫內(nèi)容一、實(shí)驗(yàn)環(huán)境二、實(shí)驗(yàn)?zāi)康娜?、?shí)驗(yàn)內(nèi)容四、數(shù)據(jù)結(jié)構(gòu)與算法思想描述五、程序清單六、程序執(zhí)行結(jié)果及其分析計(jì)算機(jī)科學(xué)與技術(shù)系2016年制一、 實(shí)驗(yàn)環(huán)境32位操作系統(tǒng)下的Window平臺 Microsoft Visual C+二、 實(shí)驗(yàn)?zāi)康恼莆諚５膶?shí)現(xiàn)及使用三、 實(shí)驗(yàn)內(nèi)容1.實(shí)現(xiàn)棧的存儲結(jié)構(gòu)2.實(shí)現(xiàn)棧的基本操作的有關(guān)算法3.利用棧解決*算術(shù)表達(dá)式求值問題四、 數(shù)據(jù)結(jié)構(gòu)與算法思想描述順序讀取中綴表達(dá)式：1、 當(dāng)遇到數(shù)字時(shí)，將數(shù)字入數(shù)字棧2、 當(dāng)遇到操作符時(shí)，與操作符棧

2、棧頂比較：If(當(dāng)前操作符優(yōu)先級大于操作符棧棧頂?shù)膬?yōu)先級)If(非”)”操作符)將當(dāng)前操作符進(jìn)操作符棧;ElseWhile(操作符棧棧頂不等于”(“)取操作符棧棧頂及數(shù)字棧的兩個(gè)數(shù)進(jìn)行運(yùn)算，并將結(jié)果壓入數(shù)字棧;ElseIf(非(“操作符)While(操作符棧棧頂不等于”(“)取操作符棧棧頂及數(shù)字棧的兩個(gè)數(shù)進(jìn)行運(yùn)算，并將結(jié)果壓入數(shù)字棧;Continue;(直到當(dāng)前操作符比棧頂操作符優(yōu)先級大)Else將當(dāng)前操作符進(jìn)操作符棧;3、 While(操作符棧非空)操作符棧棧頂及數(shù)字棧的兩個(gè)數(shù)進(jìn)行運(yùn)算，并將結(jié)果壓入數(shù)字棧;4、 在數(shù)字棧取最后結(jié)果并輸出。五、 程序清單/10*82+16.3+5*(5.2*5

3、+3.01)/4-(-10)+0.1000060+4.00416-40 = 666.666666/100+(-100)-(-102) = 100/(2016-2017+(2015-2014) = 0/-1+(10)+100%102 = 0#include<iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<iomanip>#include<map>using namespace std;const int MAX = 105;typede

4、f double Type;typedef structType TypeStackMAX;char charStackMAX;int TypeTop, charTop;Stack;/初始化棧void InitStack(Stack *S)S->charTop = S->TypeTop = 0;/判斷charStack是否為空bool IsEmpty_Char(Stack S)return S.charTop = 0;/判斷TypeStack是否為空bool IsEmpty_Type(Stack S)return S.TypeTop = 0;/判斷charStack是否為滿bool

5、 IsFull_Char(Stack S)return S.charTop = MAX;/判斷TypeStack是否為滿bool IsFull_Type(Stack S)return S.TypeTop = MAX;void Push_Char(Stack *S, char ch)/charStack不為滿則入棧，否則輸出提示if(!IsFull_Char(*S)S->charStackS->charTop+ = ch;elsecout << " The CharStack Is Full! " << endl;void Push_Typ

6、e(Stack *S, Type a)/TypeStack不為滿則入棧，否則輸出提示if(!IsFull_Type(*S)S->TypeStackS->TypeTop+ = a;elsecout << " The TypeStack Is Full! " << endl;char Pop_Char(Stack *S)if(!IsEmpty_Char(*S)S->charTop-;return S->charStackS->charTop;elsecout << " The CharStack Is

7、Empty! " << endl;return -1;Type Pop_Type(Stack *S)if(!IsEmpty_Type(*S)S->TypeTop-;return S->TypeStackS->TypeTop;elsecout << " The TypeStack Is Empty! " << endl;return -1;char Top_Char(Stack S)if(!IsEmpty_Char(S)return S.charStack-S.charTop;elsecout <<

8、" 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 '+':

9、value = left + right;break;case '-':value = left - right;break;case '*':value = left * right;break;case '/':if(right != 0)value = left / right;elsecout << "被除數(shù)不能為零!" << endl;break;case '%':if(right != 0)value = (int)left % (int)right;elsecout <

10、;< "被余數(shù)不能為零!" << endl;break;case '':value = pow(left,right);/*value = 1;if(right >= 0)while(right-)value *= left;elseright = -right;while(right-)value /= left;*/return value;void Computer(char *mid_equotion, Type len)Type right, left , result;char *p_mid_equotion = mid_

11、equotion;char after_equotion = ' 'map<char,Type> Oper;Oper'#' = 1;Oper'(' = 2;Oper'+' = 3;Oper'-' = 3;Oper'*' = 4;Oper'/' = 4;Oper'%' = 4;Oper'' = 5;Oper')' = 6;Stack MyStack;InitStack(&MyStack);Push_Char(&

12、MyStack,'#');char top_oper, current_oper;for(;*p_mid_equotion != '0')top_oper = Top_Char(MyStack);current_oper = *p_mid_equotion;if(!Opercurrent_oper)Push_Type(&MyStack,strtod(p_mid_equotion, &p_mid_equotion);continue;/end ifelse/為操作符if(Opercurrent_oper > Opertop_oper)if(c

13、urrent_oper != ')')Push_Char(&MyStack,current_oper);elsewhile(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_C

14、har(MyStack);Pop_Char(&MyStack);/end else/end ifelseif(current_oper = '(')Push_Char(&MyStack,current_oper);if(*(p_mid_equotion + 1) = '-')Push_Type(&MyStack,0);elseright = Pop_Type(&MyStack);if(!IsEmpty_Type(MyStack)left = Pop_Type(&MyStack);elseleft = 0;Push_Type

15、(&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(&M

16、yStack,Calculate(left, right, top_oper);top_oper = Pop_Char(&MyStack);/cout << setprecision(6) << "nThe Result = " << (result = Pop_Type(&MyStack) << endl;printf("The Result = %lfnn",(result = Pop_Type(&MyStack);int main()char sMAX = ""Type i = 0;cout << "請輸入你要求值的表達(dá)式!