SDUT 小型Basic编译器问题

http://acm.sdut.edu.cn/web/showproblem.php?pid=2099&cid=1088

才开始掉下了中间可能有多个空格的处理，WA了好几次。。。还有这里对表达式的理解一定要透彻，表达式有三种1：常数；2：A+B形式；3：A>B的形式。。。

#include <cstdio>
#include <cstring>
#include <iostream>
#define maxn 107
using namespace std;
struct node
{
    int num;//存编号
    char op[10];//存操作符
    char exp[10];//存表达式
    int expl;//存表达式的长度
}tagp[maxn];
int en[30];
int main()
{
    char st[maxn];
    int i,j,len,jj;
    int k = 1;
    memset(tagp,0,sizeof(tagp));
    memset(en,0,sizeof(en));
    while (gets(st) != NULL)
    {
        len = strlen(st);
        i = 0;
        while (st[i] == ' ') i++;//处理空格
        for (; st[i] != ' ' && i < len; ++i)
        {
            tagp[k].num =  tagp[k].num*10 + st[i] - '0';
        }
        while (st[i] == ' ') i++;//处理空格
        int l = 0;
        for (j = i; st[j] != ' ' && j < len; ++j)
        tagp[k].op[l++] = st[j];
        while (st[j] == ' ') j++;
        l = 0;
        for (jj = j; jj < len; ++jj)
        {
            tagp[k].exp[l++] = st[jj];
            tagp[k].expl++;
        }
        k++;
    }
    int s = 1;
    int pos1 = 0;
    int pos2 = 0;
    int pos3 = 0;
    while (s)
    {
        if (strcmp(tagp[s].op,"STOP") == 0) break;
        if (strcmp(tagp[s].op,"LET") == 0)
        {
            pos1 = tagp[s].exp[0] - 'A';
           if (tagp[s].expl > 3)//表达式长度大于3可能常数可能是其他两种
           {
               if (tagp[s].exp[3] >= '0' && tagp[s].exp[3] <= '9')//是常数
               {
                   int sum = 0;
                   for (i = 2; i < tagp[s].expl; ++i)
                   {
                       sum = sum*10 + (tagp[s].exp[i] - '0');
                   }
                   en[pos1] = sum;
               }
               else//是A>B或A+B
               {
                   pos2 = tagp[s].exp[2] - 'A';
                   pos3 = tagp[s].exp[4] - 'A';
                   if (tagp[s].exp[3] == '+')
                   en[pos1] = en[pos2] + en[pos3];
                   else if (tagp[s].exp[3] == '>')
                   {
                       if (en[pos2] > en[pos3])
                       en[pos1] = 1;
                       else
                       en[pos1] = 0;
                   }
               }
           }
           else//小于3肯定是常数
           {
                int sum = 0;
                for (i = 2; i < tagp[s].expl; ++i)
                {
                    sum = sum*10 + (tagp[s].exp[i] - '0');
                }
                en[pos1] = sum;
           }
           s++;
        }
        else if (strcmp(tagp[s].op,"PRINT") == 0)
        {
            pos1 = tagp[s].exp[0] - 'A';
            printf("%s=%d\n",tagp[s].exp,en[pos1]);
            s++;
        }
        else if (strcmp(tagp[s].op,"IF") == 0)
        {
            if (tagp[s].expl == 3)//三种情况都有
            {
               if (tagp[s].exp[1] >= '0' && tagp[s].exp[1] <= '9')//常数
               {
                   int sum = 0;
                   for (i = 0; i < tagp[s].expl; ++i)
                   {
                       sum = sum*10 + (tagp[s].exp[i] - '0');
                   }
                   if (sum > 0) s++;
                   else s += 2;
               }
               else
               {
                   pos2 = tagp[s].exp[0] - 'A';
                   pos3 = tagp[s].exp[2] - 'A';
                   if (tagp[s].exp[1] == '+')//A+B
                   {
                       int mark = en[pos2] + en[pos3];
                       if (mark > 0) s++;
                       else s += 2;
                   }
                   else if (tagp[s].exp[1] == '>')//Ａ＞Ｂ
                   {
                        pos2 = tagp[s].exp[0] - 'A';
                        pos3 = tagp[s].exp[2] - 'A';
                        if (en[pos2] > en[pos3])
                        s++;
                        else
                        s += 2;
                   }
               }
            }
            else//常数
            {
                    int sum = 0;
                    for (i = 0; i < tagp[s].expl; ++i)
                    {
                        sum = sum*10 + (tagp[s].exp[i] - '0');
                    }
                    if (sum > 0)
                    s++;
                    else
                    s += 2;
            }
        }
        //同上
        else if (strcmp(tagp[s].op,"GOTO") == 0)
        {
            if (tagp[s].expl == 3)
            {
               if (tagp[s].exp[1] >= '0' && tagp[s].exp[1] <= '9')
               {
                   int sum = 0;
                   for (i = 0; i < tagp[s].expl; ++i)
                   {
                       sum = sum*10 + (tagp[s].exp[i] - '0');
                   }
                   if (sum > 0) s = sum;
                   else break;
               }
               else
               {
                   pos2 = tagp[s].exp[0] - 'A';
                   pos3 = tagp[s].exp[2] - 'A';
                   if (tagp[s].exp[1] == '+')
                   {
                       int mark = en[pos2] + en[pos3];
                       if (mark > 0) s = mark;
                       else break;
                   }
                   else if (tagp[s].exp[1] == '>')
                   {
                        pos2 = tagp[s].exp[0] - 'A';
                        pos3 = tagp[s].exp[2] - 'A';
                        if (en[pos2] > en[pos3])
                        s = 1;
                        else
                        break;
                   }
               }
            }
            else
            {
                    int sum = 0;
                    for (i = 0; i < tagp[s].expl; ++i)
                    {
                        sum = sum*10 + (tagp[s].exp[i] - '0');
                    }
                    if (sum > 0)
                    s = sum;
                    else
                    break;
            }
        }
    }
    return 0;
}