NYOJ-35-表达式求值

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int map[7][7]=    //算符间的优先关系,100表示不会出现的情况
{
{1,1,-1,-1,-1,1,1},
{1,1,-1,-1,-1,1,1},
{1,1,1,1,-1,1,1},
{1,1,1,1,-1,1,1},
{-1,-1,-1,-1,-1,0,100},
{1,1,1,1,100,1,1},
{-1,-1,-1,-1,-1,100,0}
};
int cam(char c)
{
switch(c)
{
case '+':return 0;
case '-':return 1;
case '*':return 2;
case '/':return 3;
case '(':return 4;
case ')':return 5;
case '#':return 6;
}
}
double sol(double x,char c,double y)
{
switch(c)
{
case '+':return x+y;
case '-':return x-y;
case '*':return x*y;
case '/':return x/y;
}
}
int z(char c)
{
if('0'<=c&&c<='9'||c=='.')
return 1;
if(c==' ')
return -1;
return 0;
}
char str[1005];
char optr[1005];
double opnd[1005];
int main()
{
int t1,t2,k,len;
char ch,zz;
int temp1,temp2;
double a,b;
int t;
scanf("%d",&t);
getchar();
while(t--)
{
gets(str);
len=strlen(str);
str[len-1]='#'; //处理的等于号
t1=t2=k=0;
optr[t1++]='#';
ch=str[k++];
while(ch!='#'||optr[t1-1]!='#')
{
if(z(ch)==1)  //操作数入栈
{
opnd[t2++]=atof(&str[k-1]); //把字符串转换成浮点数
while(z(str[k])==1)
k++;
ch=str[k++];
}
else if(z(ch)==-1)
ch=str[k++];
else
{
temp1=cam(optr[t1-1]);
temp2=cam(ch);
if(map[temp1][temp2]==-1)  //栈顶元素优先权低
{
optr[t1++]=ch;
ch=str[k++];
}
else if(map[temp1][temp2]==0) //脱括号并接受下一个字符
{
t1--;
ch=str[k++];
}
else    //退栈并将运算结果
{
zz=optr[--t1];
a=opnd[--t2];
b=opnd[--t2];
opnd[t2++]=sol(b,zz,a);
}
}
}
printf("%.2lf\n",opnd[0]);
}
return 0;
}