jisuanqi.rar_计算器

/* 说明: 采用树形结构处理表达式,按优先级运算结果,一个加,减,乘,除或数值为一个节点 优先级如下: 函数:4 括号:3 乘除:2 加减:1 */ #include <windows.h> #include <iostream> #include <fstream> #include <string> #include <cmath> using namespace std; const char NUM[]={'0','1','2','3','4','5','6','7','8','9','.'}; const char OPERATION[]={'+','-','*','/'}; const double PI=3.14159265358979; const double EE=2.71828182818281; class Fun //处理系统数学函数的类 { public: Fun(string o,int t,double l=0.0,double r=0.0):op(o),type(t),lvalue(l),rvalue(r){} static string FUN[]; double calc(); private: int type; //666 0 1 sin90 2 3! 3 3C2 string op; //函数类型 double lvalue; //函数左边的值 double rvalue; //函数右边的值 static int FunNum; }; int Fun::FunNum=10; string Fun::FUN[]={"!","sin","cos","tan","log","ln","C","A","^","-"}; /* 函数说明: 1:log是以10为底的工程对数 2:ln 是以e为底的自然对数 3:C 计算组合数 输入规则 如计算 3取2的组合 输入表达式 3C2 4:A 计算排列数 输入规则 如计算 3取2的排列 输入表达式 3A2 5:! 计算阶乘 6:^ x的y次方 输入 x^y */ int factorial(int n) //阶乘函数 { int i,s=1; for(i=1;i<=n;i++) s*=i; return s; } int C(int a,int b) { return factorial(a)/(factorial(b)*factorial(a-b)); } int A(int a,int b) { return factorial(a)/factorial(b); } double Fun::calc() //计算系统函数的值 { if(type==0) return lvalue; else { if(op=="!") return factorial(lvalue); if(op=="sin") return sin(rvalue/180*PI); if(op=="cos") return cos(rvalue/180*PI); if(op=="tan") return tan(rvalue/180*PI); if(op=="log") return log10(rvalue); if(op=="ln") return log10(rvalue)/log10(EE); if(op=="C") return C(lvalue,rvalue); if(op=="A") return A(lvalue,rvalue); if(op=="^") return pow(lvalue,rvalue); if(op=="-") return -rvalue; else { string err="暂时没有函数"+op; MessageBox(NULL,err.c_str(),"错误",MB_OK); return 0; } } } struct Unit //双向链表保存运算单元 { Unit(int p,char o,string c,double v,int t,Unit * pr=NULL,Unit * n=NULL) :PRI(p),Operation(o),Code(c),value(v),Type(t),Pre(pr),Next(n){} int PRI; //优先级 char Operation; //操作符 string Code; //原始代码 double value; //数据 int Type; //类型 操作符0 数据1 函数2 Unit * Pre; //构成双向链表 Unit * Next; }; class Node //表达式树状结构的节点 { public: Node(char o,int p,int e=1,double v=0,Node * ph=NULL,Node * pl=NULL,Node * pr=NULL) :Operation(o),PRI(p),Expression(e),value(v),Head(ph),Left(pl),Right(pr){} Node * Head; //节点的根,左树枝,右树枝 Node * Left; Node * Right; double GetValue(); char GetOperation() const {return Operation;} int GetPri() const {return PRI;} int IsExp() const {return Expression;} private: char Operation; //操作符 int PRI; //优先级 int Expression; //记录该节点是否是表达式0 1 double value; //该节点的值 }; double Node::GetValue() //运算该节点的值 { if(IsExp()) //该节点的值还未算出来 { double lvalue,rvalue; lvalue=Left->GetValue(); rvalue=Right->GetValue(); Expression=0; char op=GetOperation(); switch(op) { case '+': return lvalue+rvalue; case '-': return lvalue-rvalue; case '*': return lvalue*rvalue; case '/': return lvalue/rvalue; default: return 0; } } else return value; } bool Isnum(char c) { for(int i=0;i<sizeof(NUM);i++) { if(c==NUM[i]) return true; } return false; } bool Isoperation(char c) { for(int i=0;i<sizeof(OPERATION);i++) { if(c==OPERATION[i]) return true; } return false; } Unit * Analyse(string exp) //分析表达式并生成链表 { int pri=0; //当前优先级 int stat=1; //当前的读入状态 括号 0 运算符 1 其他 2 Unit * head=NULL,* p=NULL; int i=0,explen; explen=exp.size(); for(i=0;i<explen;i++) { char c=exp.at(i); if(c=='(') { pri+=3; stat=0; } else if(c==')') { pri-=3; stat=0; } else if(Isoperation(c) && stat!=1) //操作符后的当正负号处理 { stat=1; Unit * temp=p; int add_pri; //自身增加的优先级 if(c=='+' || c=='-') add_pri=1; else add_pri=2; p->Next=new Unit(pri+add_pri,c," ",0,0); p=p->Next; p->Pre=temp; } else //其他的当做函数处理 { stat=2; string function=""; while(i<explen && (c=exp.at(i),! Isoperation(c)) && c!=')') { function+=c; i++; } i--; if(head==NULL) { p=new Unit(pri,' ',function,0,2); head=p; } else { Unit * temp=p; p->Next=new Unit(pri,' ',function,0,2); p=p->Next; p->Pre=temp; } } } return head; } Unit * Calc(Unit * head) //计算双向链表基本单元的值 { Unit * p=head; while(p!=NULL) { if(p->Type!=0) //非操作符 { string temp=p->Code; string op; double lvalue=0,rvalue=0; int l_point=0,r_point=0; int i=0,type=0; char ch; while(i<temp.size() && (ch=temp.at(i),Isnum(ch))) { if(ch=='.') { l_point++; i++; continue; } if(! l_point) lvalue*=10; lvalue+=(ch-'0')*pow(10,-l_point); i++; if(l_point) l_point++; } while(i<temp.size() && (ch=temp.at(i),! Isnum(ch))) { op+=ch; type=1; i++; } while(i<temp.size() && (ch=temp.at(i),Isnum(ch))) { if(ch=='.') { r_point++; i++; continue; } if(! r_point) rvalue*=10; rvalue+=(ch-'0')*pow(10,-r_point); i++; if(r_point) r_point++; } Fun * f=new Fun(op,type,lvalue,rvalue); p->value=f->calc(); } p=p->Next; } return head; } Node * Tree(Unit * head) //生成表达式树 { Node * root=NULL,* proot=NULL,* pbranch=NULL; Unit * p=head; int now_pri; //当前优先级 bool hadop=false; while(p!=NULL) { if(p->Type==0) //如果是一个操作符 { hadop=true; if(root==NULL) { proot=new Node(p->Operation,p->PRI,1); root=proot; pbranch=root; now_pri=p->PRI; proot->Left=new Node(' ',0,0,p->Pre->value); proot->Right=new Node(' ',0,0,p->Next->value); } else { if(p->PRI<now_pri) //优先级低于当前优先级,树根方向 { proot=new Node(p->Operation,p->PRI,1); //新的树根 proot->Left=root; //根的变换 proot->Left->Head=proot; proot->Right=new Node(' ',0,0,p->Next->value); proot->Right->Head=proot; root=proot; pbranch=proot; //右树枝的变换 //pbranch->Right=new Node(' ',0,0,p->Pre->value); //树枝右边取值 } else if(p->PRI==now_pri) //优先级相同,先算左边的 { Node * temp; temp=new Node(p->Operation,p->PRI,1); temp->Left=pbranch; if(pbranch->Head==NULL) { proot=temp; root=proot; pbranch->Head=proot; } else { Node * temp0; temp0=pbranch->Head; temp0->Right=temp; temp->Head=temp0; pbranch->Head=temp; } pbranch=temp; pbranch->Right=new Node(' ',0,0,p->Next->value); pbranch->Right->Head=pbranch; } else { Node * temp; temp=new Node(p->Operation,p->PRI,1); pbranch->Right=temp; temp->Head=pbranch; pbranch=pbranch->Right; pbranch->Left=new Node(' ',0,0,p->Pre->value); pbranch->Left->Head=pbranch; pbranch->Right=new Node(' ',0,0,p->Next->value); pbranch->Right->Head=pbranch; } now_pri=p->PRI; } } p=p->Next; } if(! hadop) root=new Node(' ',0,0,head->value); return root; } int main() { string exp; ifstream infile("test.txt",ios::in); while(! getline(infile,exp).eof()) { if(exp=="") continue; Unit * h=Analyse(exp); h=Calc(h); Node * root=Tree(h); cout<<exp<<"="<<root->GetValue()<<endl; } return 0; } 备注:修正了上一篇的一个bug,支持负数运算. (((3+2)*9)-(6/3)*sin90+1)/8+18*3=59.5 6.6+3.2*6=25.8 sin60+3*2=6.86603 3+2*6/3+12/6=9 sin30*2+3*6-sin90*(6+3)=10 1=1 1+2=3 4C2+4A2*3=42 3^3/9-9*6=-51 tan45+sin30=1.5 9/3/3=1 9/3/3/3=0.333333 1+1+2+3*3=13 1*2+3=5 2*3+3*6=24 3+3+3+3+3=15 3*3*3+3*3*3=54 -1=-1 -1+-2=-3 -1*(-2+3)*-3=-3 Press any key to continue