描述：思路很简单，优先级比较完的情况下，采用两个栈就可以完成了……
#include <iostream>#include <cstdio>#include <cstring>#include "hanshu.h" //头文件using namespace std;int flag;//下面会多次用到flagint num_value(char *s,char *str,double *num,int len,int i,int j);//计算实数值存入num中char get_precede(char a,char b);//比较两个字符优先级，获取优先级int calculate_num(char c,double *num,int j);//优先级比较大小后计算结果double solve_s(char *s ,char *str,double *num,int len);//处理函数，得出等式的结果int main(){ freopen("a.txt","r",stdin); double num[10010]; char c,s[10010],str[10010]; memset(s,0,sizeof(s)); int left,right,len; left=right=len=flag=0; while(scanf("%c",&c)!=EOF)//单字符输入 { if((c>='0'&&c<='9')||c=='.'||c=='('||c==')'||c=='+'||c=='-'||c=='*'||c=='/')//选择合适的字符输入 { s[len++]=c; if(c=='(') { if(!right)flag=1; left++; } else if(c==')') { right++; if(!left) flag=2; } } if(c=='\n')//讨论几种不符合情况的输入 { if(flag==2) printf("输入错误，首先出现了右括号！\n"); else if(left>right) printf("左括号比右括号多!\n"); else if(left<right) printf("右括号比左括号多!\n"); else if(len<=0) printf("没有数据!\n"); else { s[len++]='#'; flag=0; memset(str,0,sizeof(str)); memset(num,0,sizeof(num)); double x=solve_s(s,str,num,len); if(flag) printf("出现零为除数\n"); else printf("solve_s=%g\n",x); } memset(s,0,sizeof(s)); flag=len=left=right=0; } } return 0;}//头文件
#ifndef HANSHU_H_INCLUDED#define HANSHU_H_INCLUDEDchar str_operator[]= {'+','-','*','/','(',')','#'};char str_precede[7][7]={ {'>','>','<','<','<','>','>'}, {'>','>','<','<','<','>','>'}, {'>','>','>','>','<','>','>'}, {'>','>','>','>','<','>','>'}, {'<','<','<','<','<','=',' '}, {'>','>','>','>',' ','>','>'}, {'<','<','<','<','<',' ','='}};extern int flag;int num_value(char *s,char *str,double *num,int len,int i,int j){ int k,count(0),c(0); for( k=i; k<len; k++) if(s[k]>='0'&&s[k]<='9') { num[j]=num[j]*10+s[k]-'0'; if(c) count++; } else if(s[k]=='.') c=1; else break; for( c=0; c<count; c++) num[j]/=10; if((i==1&&(s[i-1]=='+'||s[i-1]=='-'))||(i>=2&&(s[i-1]=='-'||s[i-1]=='+')&&s[i-2]=='(')) { flag=1; if(s[i-1]=='-') num[j]=0-num[j]; } return k;}char get_precede(char a,char b){ int i,j; for(int k=0; k<7; k++) { if(a==str_operator[k]) i=k; if(b==str_operator[k]) j=k; } return str_precede[i][j];}int calculate_num(char c,double *num,int j){ j--; switch(c) { case '+': num[j-1]=num[j-1]+num[j]; break; case '-': num[j-1]=num[j-1]-num[j]; break; case '/': num[j-1]=num[j-1]/num[j]; break; case '*': num[j-1]=num[j-1]*num[j]; break; } if(c=='-'||c=='+'||c=='*'||c=='/') { num[j]=0; flag=1; return j; } else return j+1;}double solve_s(char *s ,char *str,double *num,int len){ int i,j,k(1); str[0]='#'; for(i=j=0; i<len; i++) { if(s[i]>='0'&&s[i]<='9') { i=num_value(s,str,num,len,i,j); if(flag) { str[--k]=0; flag=0; } j++; } char c=get_precede(str[k-1],s[i]); if(c=='<') str[k++]=s[i]; else if(c=='=') str[--k]=0; else if(c=='>') { while(c=='>') { if(str[k-1]=='/'&&!num[j-1]) { flag=1; return -1; } flag=0; j=calculate_num(str[k-1],num,j); if(flag) { str[--k]=0; c=get_precede(str[k-1],s[i]); flag=0; continue; } else break; } if(str[k-1]=='('&&s[i]==')') str[--k]=0; else str[k++]=s[i]; } } return num[0];}#endif // HANSHU_H_INCLUDED
