NYOJ 35题 表达式求值，四则运算（栈实现） 严蔚敏

这些函数中都有“重复”的，因为操作数(OPND)栈用double，操作符(OPTR)栈用char。C++中的模板可以解决这个问题吗？

 这是对着书写的：

#include <iostream> using namespace std; #define STACK_INIT_SIZE 100 #define STACKINCREMENT 100 char Precede_Matrix[7][7] = { {'>', '>', '<', '<', '<', '>', '>',}, {'>', '>', '<', '<', '<', '>', '>',}, {'>', '>', '>', '>', '<', '>', '>',}, {'>', '>', '>', '>', '<', '>', '>',}, {'<', '<', '<', '<', '<', '=', '0',}, {'>', '>', '>', '>', '0', '>', '>',}, {'<', '<', '<', '<', '<', '0', '=',} }; char Precede ( char a, char b ) { int i = 0; int j = 0; switch (a) { case '+' : i = a - '+'; break; case '-' : i = a - '-' + 1; break; case '*' : i = a - '*' + 2; break; case '/' : i = a - '/' + 3; break; case '(' : i = a - '(' + 4; break; case ')' : i = a - ')' + 5; break; case '#' : i = a - '#' + 6; break; default : cout << "Error1!" << endl; } switch (b) { case '+' : j = b - '+'; break; case '-' : j = b - '-' + 1; break; case '*' : j = b - '*' + 2; break; case '/' : j = b - '/' + 3; break; case '(' : j = b - '(' + 4; break; case ')' : j = b - ')' + 5; break; case '#' : j = b - '#' + 6; break; default : cout << "Error2!" << endl; } char c = Precede_Matrix[i][j]; return c; } int Operate(int a, char oper, int b) { switch (oper) { case '+': return a + b; case '-': return a - b; case '*': return a * b; case '/': return a / b; default: cout << "Error3!" << endl; return 0; } } struct stack_char { char *base; char *top; int stacksize; }OPTR; struct stack_int { int *base; int *top; int stacksize; }OPND; // 两位数的只能用int来存？ void InitStack(struct stack_char &S) { S.base = (char *)malloc( STACK_INIT_SIZE * sizeof(char) ); if(!S.base) cout << "Overflow1!" << endl; S.top = S.base; S.stacksize = STACK_INIT_SIZE; } void InitStack(stack_int &S) { S.base = (int *)malloc( STACK_INIT_SIZE * sizeof(int) ); if(!S.base) cout << "Overflow2!" << endl; S.top = S.base; S.stacksize = STACK_INIT_SIZE; } void Push(stack_char &S, char e) { if(S.top - S.base >= S.stacksize) // 判满 { S.base = (char *)realloc( S.base, (S.stacksize + STACKINCREMENT) * sizeof(char) ); if(!S.base) cout << "Overflow3!" << endl; S.top = S.base + S.stacksize; S.stacksize += STACKINCREMENT; } * S.top++ = e; } void Push(stack_int &S, int e) { if(S.top - S.base >= S.stacksize) // 判满 { S.base = (int *)realloc( S.base, (S.stacksize + STACKINCREMENT) * sizeof(int) ); if(!S.base) cout << "Overflow4!" << endl; S.top = S.base + S.stacksize; S.stacksize += STACKINCREMENT; } * S.top++ = e; } void Pop (stack_char &S, char &e) { if(S.top == S.base) // 判空 cout << "Error4!" << endl; e = * --S.top; } void Pop (stack_int &S, int &e) { if(S.top == S.base) // 判空 cout << "Error5!" << endl; e = * --S.top; } char GetTop(stack_char S) { if (S.top == S.base) // 判空 cout << "Error6!" << endl; return *(S.top - 1); } int GetTop(stack_int S) { if (S.top == S.base) // 判空 cout << "Error7!" << endl; return *(S.top - 1); } int EvaluateExpression() { int a; int b; char x; char theta; InitStack(OPTR); Push(OPTR, '#'); InitStack(OPND); char c = getchar(); // 输入若是数字，只能是个位数 while ( !(c == '#' && GetTop(OPTR) == '#') ) { if ( '0' <= c && c <= '9' ) // c is operand { c -= '0'; Push(OPND, c); c = getchar(); } else // c is operator or delimiter { switch (Precede (GetTop(OPTR), c)) { case '<': Push(OPTR, c); c = getchar(); break; case '=': Pop(OPTR, x); c = getchar(); break; case '>': Pop(OPTR, theta); Pop(OPND, b); Pop(OPND, a); Push(OPND, Operate(a, theta, b)); break; default: cout << "Error8!" << endl; } } } // OPTR栈的栈顶元素和当前读入的字符均为“#” // 即“#”=“#”时整个表达式求值完毕 int result = GetTop(OPND); free(OPTR.base); free(OPND.base); return result; // 要在释放前保留下结果 } int main() { cout << EvaluateExpression() << endl; return 0; }

NYOJ上该题的标程：

#include<stdio.h> #include<string.h> #include<stdlib.h> char str[1005]; int start; char s[50],ss[50];int i,j; double Term();double Expression();double Factor(); double Term() { double f=Factor(),t; --start; if(str[start]=='*') { t=Term(); return t*f; } else if(str[start]=='/') { t=Term(); return t/f; } else {++start;return f;} } double Expression() { double t=Term(),e; --start; if(str[start]=='+') { e=Expression(); return e+t; } else if(str[start]=='-') { e=Expression(); return e-t; } else {++start;return t;} } double Factor() { --start; double ret; if(str[start]==')') { ret=Expression(); --start; return ret; } else { i=0; memset(ss,0,sizeof(ss)); while((str[start]>='0'&&str[start]<='9')||str[start]=='.') s[i++]=str[start--]; ++start; for(j=0;j<i;j++) { ss[j]=s[i-1-j]; } return atof(ss); } } int main() { //freopen("1.txt","r",stdin); int n; scanf("%d",&n); while(n--) { scanf("%s",str); start=strlen(str)-1; printf("%.2lf\n",Expression()); } }

2012/5/17 更新：

终于用模板解决了那个问题

#include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> using namespace std; #define STACK_INIT_SIZE 1024 #define STACKINCREMENT 30 char Precede_Matrix[7][7] = { {'>', '>', '<', '<', '<', '>', '>',}, {'>', '>', '<', '<', '<', '>', '>',}, {'>', '>', '>', '>', '<', '>', '>',}, {'>', '>', '>', '>', '<', '>', '>',}, {'<', '<', '<', '<', '<', '=', '0',}, {'>', '>', '>', '>', '0', '>', '>',}, {'<', '<', '<', '<', '<', '0', '=',} }; char Precede ( char a, char b ) { int i = 0; int j = 0; switch (a) { case '+' : i = 0; break; case '-' : i = 1; break; case '*' : i = 2; break; case '/' : i = 3; break; case '(' : i = 4; break; case ')' : i = 5; break; case '#' : i = 6; break; default : cout << "Error1!" << endl; } switch (b) { case '+' : j = 0; break; case '-' : j = 1; break; case '*' : j = 2; break; case '/' : j = 3; break; case '(' : j = 4; break; case ')' : j = 5; break; case '#' : j = 6; break; default : cout << "Error2!" << endl; } return ( Precede_Matrix[i][j] ); } double Operate(double a, char oper, double b) { switch (oper) { case '+': return a + b; case '-': return a - b; case '*': return a * b; case '/': return a / b; default: cout << "Error3!" << endl; return -1; } } struct stack_char { char *base; char *top; int stacksize; }; struct stack_double { double *base; double *top; int stacksize; }; // 计算结果可能是两位数，只能用double，不能用char来存？ template <class T2, class T1> void InitStack(T1 &S) { S.base = (T2 *)malloc( STACK_INIT_SIZE * sizeof(T2) ); if(!S.base) cout << "Overflow2!" << endl; S.top = S.base; S.stacksize = STACK_INIT_SIZE; } template <class T1, class T2> void Push(T1 &S, T2 e) { if(S.top - S.base >= S.stacksize) // 判满 { S.base = (T2 *)realloc( S.base, (S.stacksize + STACKINCREMENT) * sizeof(T2) ); if(!S.base) cout << "Overflow3!" << endl; S.top = S.base + S.stacksize; S.stacksize += STACKINCREMENT; } *S.top++ = e; } template <class T1, class T2> void Pop (T1 &S, T2 &e) { if(S.top == S.base) // 判空 cout << "Error5!" << endl; e = *--S.top; } template <class T2, class T1> T2 GetTop(T1 S) { if (S.top == S.base) // 判空 cout << "Error6!" << endl; return *(S.top - 1); } bool IsOperand(char c) { if( ('0' <= c && c <= '9') || c == '.' ) // c是数字或小数点 return true; else return false; } int main(void) { string str; while (cin >> str) { str.push_back('#'); // 最后是#（结束标志） double a; double b; char x; char theta; stack_char OPTR; InitStack<char> (OPTR); Push(OPTR, '#'); stack_double OPND; InitStack<double> (OPND); int i = 0; char c = str[i++]; double operand = 0; while ( !(c == '#' && GetTop<char> (OPTR) == '#') ) { if ( IsOperand(c) ) // c is operand { operand = atof( &str[i - 1] ); // 把从c开头的数转化成double Push(OPND, operand); while( IsOperand(str[i]) ) i++; c = str[i++]; } else // c is operator or delimiter { switch (Precede (GetTop<char> (OPTR), c)) { case '<': Push(OPTR, c); c = str[i++]; break; case '=': Pop(OPTR, x); c = str[i++]; break; case '>': Pop(OPTR, theta); Pop(OPND, b); Pop(OPND, a); Push(OPND, Operate(a, theta, b)); break; default: cout << "Error8!" << endl; break; } } } // OPTR栈的栈顶元素和当前读入的字符均为“#” // 即“#”=“#”时整个表达式求值完毕 cout << GetTop<double> (OPND) << endl; free(OPTR.base); free(OPND.base); } return 0; }

同时功能上改进了，欢迎体验！