ACboy needs your help again!--hdu1702
ACboy needs your help again!
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4429 Accepted Submission(s): 2260
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!
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.
#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
using namespace std;
char s1[],s2[];
int main()
{
int n,a,b; scanf("%d",&n);
while(n--)
{
queue<int>s;//定义一个队列
while(!s.empty())
s.pop();
stack<int>r;//定义一个栈
while(!r.empty())
r.pop();//一定要每次都要清空队列,要不然各种WA我也是醉了!
scanf("%d %s",&a,s1);
getchar();//注意吸收回车
if(strcmp(s1,"FIFO")==)
{
while(a--)
{
scanf("%s",s2);
if(strcmp(s2,"IN")==)
{
scanf("%d",&b);
getchar();
s.push(b);
}
else if(strcmp(s2,"OUT")==)
{
if(s.empty())
printf("None\n");
else
{
printf("%d\n",s.front());//输出队首元素
s.pop();
} }
}
}
else if(strcmp(s1,"FILO")==)
{
while(a--)
{
scanf("%s",s2);
if(strcmp(s2,"IN")==)
{
scanf("%d",&b);
getchar();
r.push(b);
}
else if(strcmp(s2,"OUT")==)
{
if(r.empty())
printf("None\n");
else
{
printf("%d\n",r.top());//输出栈顶元素
r.pop();
} }
}
}
}
return ;
}
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