miss ber multiple pri use integer print present depend

題目描述

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!

輸入

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.

輸出

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.

樣例輸入

2
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT

樣例輸出

1
2
2
1
解法一://棧，隊列的模擬

//FIFO 隊列 先進先出 FILO 棧 後進先出 

#include<stdio.h>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
queue<int> q;
stack<int> s;
int n;

void Queue(){
int num;
char str[20];
while(n--){

scanf("%s", str);
if(str[0] == ‘I‘){
scanf("%d", &num);
q.push(num);
}
else if(str[0] == ‘O‘ && !q.empty()){
printf("%d\n", q.front());
q.pop();
}
else{
printf("None\n");
}
}
}
void Stack(){
int num;
char str[20];
while(n--){
scanf("%s", str);
if(str[0] == ‘I‘){
scanf("%d", &num);
s.push(num);
}
else if(str[0] == ‘O‘ && !s.empty()){
printf("%d\n", s.top());
s.pop();
}
else{
printf("None\n");
}
}
}
int main(){
int t;
char str[10];
scanf("%d", &t);
while(t--){
while(!q.empty()){
q.pop();
}

while(!s.empty()){
s.pop();
}
scanf("%d %s", &n, str);
if(str[2] == ‘F‘)
Queue();
else
Stack();
}
return 0;
}

//為啥要這麽麻飯尼?

看解法二:

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int n,m,i,k,j,a[100];
char s[5],s1[4];
cin>>n;
while(n--)
{
scanf("%d%s",&m,s);
int t=0,p=0;
memset(a,0,sizeof(a));
if(s[2]==‘F‘)
{
for(i=0;i<m;i++)
{
cin>>s1;

if(s1[0]==‘I‘)
{
scanf("%d",&k);
a[t++]=k;
}

if(s1[0]==‘O‘)
p++;

}

for(j=0;j<p;j++)
printf("%d\n",a[j]);
}


if(s[2]==‘L‘)
{
for(i=0;i<m;i++)
{
cin>>s1;
if(s1[0]==‘I‘)
{
scanf("%d",&k);
a[t++]=k;
}
if(s1[0]==‘O‘)
p++;
}
for(j=p-1;j>=0;j--)
printf("%d\n",a[j]);
}
}
return 0;
}

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