Acboy needs your help again!

ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4439    Accepted Submission(s): 2265

Problem Description

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems, you will die with ACboy."
The problems of the monster is shown on the wall:
Each problem's first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").
and the following N lines, each line is "IN M" or "OUT", (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!

 

 

Input

The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.

 

 

Output

For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.

 

 

Sample Input

   
   
   
   

    
    
    
    4
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT
5 FIFO
IN 1
IN 2
OUT
OUT
OUT
5 FILO
IN 1
IN 2
OUT
IN 3
OUT
   
   
   
   

 

 

Sample Output

   
   
   
   

    
    
    
    1
2
2
1
1
2
None
2
3
   
   
   
   

 

 

这道题有两种方法可以解决～

方法一：数组

#include<stdio.h>
#include<string.h>
int c[10000],b[10000];
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		int m,t,j,j1,j2;
		char a[5],s[5];
		scanf("%d%s",&m,a);
		if(!strcmp(a,"FILO"))
		{
			j=j1=j2=0;
			while(m--)
			{
				scanf("%s",s);
				if(!strcmp(s,"IN"))
				{
					scanf("%d",&t);
					c[j++]=t;
					j2++;
				}
				else 
				{
					if(j==0)
						printf("None\n");
					else
					{
						printf("%d\n",c[j-1]);
						j--;
					}
				}
			}
		}
		else if(!strcmp(a,"FIFO"))
		{
			j=j1=j2=0;
			while(m--)
			{
				scanf("%s",s);
				if(!strcmp(s,"IN"))
				{
					scanf("%d",&t);
					b[j++]=t;
				}
				else
				{
					if(j1>=j)
						printf("None\n");
					else
					{
						printf("%d\n",b[j1++]);
					}
				}
			}
		}
	}
	return 0;
}

方法二：栈和队列～～

#include<stdio.h>
#include<queue>
#include<string.h>
#include<stack>
#include<iostream>
using namespace std;
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{	
		char a[10];
		int m;
		scanf("%d%s",&m,a);
		if(!strcmp(a,"FIFO"))
		{
			queue<int> q;
			while(m--)
			{
				char s[10];
				scanf("%s",s);
				if(!strcmp(s,"IN"))
				{
					int b;
					scanf("%d\n",&b);
					q.push(b);
				}
				else
				{
					if(q.empty())
						printf("None\n");
					else
					{
						printf("%d\n",q.front());
						q.pop();
					}
				}	
			}
		}
		else
		{
			stack<int>w;
			while(m--)
			{
				char s[10];
				scanf("%s",s);
				if(!strcmp(s,"IN"))
				{
					int b;
					scanf("%d",&b);
					w.push(b);
				}
				else
				{
					if(w.empty())
						printf("None\n");
					else
					{
						printf("%d\n",w.top());
						w.pop();
					}
				}
			}
		}
	}
	return 0;
}