ccf - 20190302 24.2

reference

 

This question is arithmetic, we use the stack to solve it

Use template class stack c ++ in

#include <Stack>          // header 

Stack < int > NUM;    　　// define a stack type int 

num.push ();             // in the stack into a stack element 

num.pop ();              // delete the stack elements on 

num.top ();              // returns the element of the stack, and does not remove   

num.empty ();            // return stack is empty 

num.size ();             // returns the current number of elements in the stack  

1. When a character is a digit, num into the stack

2. When the character is "+", the sign is stored in the stack

3. When the character is "-" when, because of a reduced number equal to the opposite of a number of add, so that the next character is stored negated num stack, stack stores Sign "+"

4. When the character is "X", the current character to take the next character q and p, the q * p deposit num

5. When the character is "/", the current character to take the next character q and p, the q / P stores num

Code:

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
int n;
char str[10];
stack<int> num;
stack<char> sign;
int main()
{
    scanf("%d",&n);
    getchar();
    while(!num.empty()) num.pop();
    while(!sign.empty()) sign.pop();
    for(int i=0;i<n;i++)
    {
        gets(str);
        int j=0;
        while(j<strlen(str))
        {
            if(str[j]>'0'&&str[j]<='9')
            {
                num.push(str[j]-'0');
            }
            else if(str[j]=='+')
            {
                sign.push('+');
            }
            else if(str[j]=='-')
            {
                j++;
                num.push((-1)*(str[j]-'0'));
                sign.push('+');
            }
            else if(str[j]=='x')
            {
                int q=num.top();
                num.pop();
                j++;
                int p=str[j]-'0';
                num.push(q*p);
            }
            else if(str[j]=='/')
            {
                int q=num.top();
                num.pop();
                j++;
                int p=str[j]-'0';
                num.push(q/p);
            }
            j++;
        }
        while(!sign.empty())
        {
            int q=num.top();
            num.pop();
            int p=num.top();
            num.pop();
            sign.pop();
            num.push(q+p);
        }
        //printf("%d\n",num.top());
            if(num.top()==24) printf("Yes\n");
        else printf("No\n");
    }
    
    
    return 0;
}