编程作业-我的解法

来源：互联网发布：在线预约系统asp源码编辑：程序博客网时间：2024/05/18 04:54

都是上面那个编程作业我的思路是用栈进行逆序执行然后就能不用递归做出来了

#include<iostream>

#include<stack>
#include<string>
#include<cctype>
#include<sstream>
#include <strstream>
#define l_num 22
using namespace std;
bool isNum(string str)
{
   string num = "1234567890";
   if (str.find_first_not_of(num) == string::npos)
       return true;
   return false;
}
string inttow(int n) {
   string str = "";
   while (n)
   {
       string temp(1, n % 10 + '0');
       str = temp + str;
       n -= n % 10;
       n =int(n/10.0);
   }
   return str;
}
int main()
{
   int num;
   int i;
   cin >> num;
   string s[20];//本来是动态分配的
   string s1;
   int n, x;
   string b;
   for (i = 0; i < num; i++)
   {
       cin >> s[i];
   }

   do
   {
       cin >> b;
       if (b == "over") { break; }
       else
       {//如果没有结束
           string a[l_num];
           a[0] = b;
           i = 1;
           if (a[0] == "printall")
           {
               int cal;
               for (cal = 0; cal < num; cal++)
               {
                   cout << s[cal];
                   cout << endl;
               }

           }
           else
       {
   do
       {
      cin >> a[i];
      i++;
         } while (getchar()!='\n');
       stack<int> st;
       stack<string> ss;
       for (i--; i >= 0; i--)
       {
           int temp;
           if (isdigit(a[i][0]))//如果是数字进行转换并压栈
           {
               temp = atoi(a[i].c_str());//进行数字转换并压栈
               st.push(temp);
           }
           else
           {
           if (a[i] == "find")
           {
                   x = st.top();   st.pop();
                  n = st.top();st.pop();
                   stringstream sd;
                      sd << x;
                   string str = sd.str();
                   int k = s[n - 1].find(str);//改了 0
                   k = (k == string::npos ? s[n - 1].length() : k);
                   st.push(k);//执行find 的功能
           }
           else if (a[i] == "rfind")
           {
               int s2, n;
               s2 = st.top();
               st.pop();
               n = st.top();
               st.pop();
               stringstream sd;
                   sd << s2;
                   string str = sd.str();
                   int k = s[n - 1].rfind(str);
                   k = (k == string::npos ? s[n - 1].length() : k);
               st.push(k);//执行find 的功
           }
           else if (a[i] == "add")
           {
               string s2;
               string c;
               s1 = ss.top(); ss.pop();
               s2= ss.top(); ss.pop();
               if (isNum(s1) && isNum(s2))
               {
                   int a = atoi(s1.c_str());
                   int b = atoi(s2.c_str());
                   char aa[100];
                   //10是十进制的意思
                   //openjudge上面，不能使用itoa和sprintf
                   //          itoa(a+b, aa, 10);
                   //          sprintf(aa, "%d", a+b);
                   strstream ss;
                   int sum = a + b;
                   ss << sum;
                   ss >> c;
               }
               else
                   c = s1 + s2;
               ss.push(c);//执行find 的功

           }
           else if (a[i] == "copy")
           {
               string t;
               int n,x, l;
               n = st.top(); st.pop();
               x = st.top(); st.pop();
               l = st.top();   st.pop();
                 t = s[n - 1].substr(x, l);
               ss.push(t);//执行find 的功
           }
           else if (a[i] == "reset")
           {//将第n个字符串变为s
              string s2 = ss.top(); ss.pop();
              int n= st.top(); st.pop();
              s[n - 1] = s2;
           }
           else if (a[i] == "insert")
           {
           s1 = ss.top(); ss.top();
               int n, x;
               n = st.top(); st.pop();
               x= st.top(); st.pop();
               s[n - 1].insert(x, s1);
           }
           else if (a[i] == "print")
           {
               int n = st.top(); st.pop();
               cout << s[n - 1];
               cout << endl;
           }

           else
           {
               ss.push(a[i]);
           }
           }
       }
           }

       }

   } while (b != "over");

   return 0;
}

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