poj1686Lazy Math Instructor

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http://poj.org/problem?id=1686
构造栈，表达式求值。我只会C语言。。
#include <stdio.h>#include <string.h>char opp[8] = {'+','-','*','/','(',')','#'}; int priority[7][7] = {{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,-2},{1,1,1,1,-2,1,1},{-1,-1,-1,-1,-1,-2,0}};//算符之间的优先关系int calculate(int x,int y, char op) {    switch(op)    {        case '+':return x+y;        case '-':return x-y;        case '*':return x*y;        case '/':return x/y;    }}int Compare(char a,char b)//比较优先级{    int row,col,i;    for(i=0; i<7; i++)    {        if( a == opp[i] )           row = i;        if( b == opp[i] )           col = i;    }    return priority[row][col];}int result(char str[],int plus)//求表达式的结果{    char ch, op, opetator[100];    int x,y,i=0,count1=0,count2=0,operand[100];    opetator[count2++] = '#';    while(str[i]!='#' || opetator[count2-1]!='#')    {        ch = str[i];        if(ch>='0'&&ch<='9')//阿拉伯数字‘0’～‘9’        {            operand[count1++] = ch-'0';            i++;            continue;        }        if((ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z'))//英文字母        {            operand[count1++] = ch-'a'+plus;            i++;            continue;        }        if(ch=='+' || ch=='-' || ch=='*' || ch=='(' || ch==')' || ch=='#')//操作符        {            switch(Compare(opetator[count2-1],ch))            {                case -1: opetator[count2++] = ch; i++; break;//当前元素大于栈顶元素                case 0 : count2--; i++; break;//当前元素等于栈顶元素                case 1 : op = opetator[--count2]; y = operand[--count1]; x = operand[--count1];                             operand[count1++] = calculate(x,y,op); break;//当前元素小于栈顶元素            }        }        else            i++;    }    return operand[count1-1];}int main(){    char fir[90], sec[90];    int N;    scanf("%d",&N);    getchar();    while(N--)    {        gets(fir);        gets(sec);        fir[strlen( fir )] = '#';        sec[strlen( sec )] = '#';        if(result(fir,10) != result(sec,10))            printf("NO\n");          else            printf("YES\n");    }    return 0;}
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