LightOJ 1212 deque的运用

来源：http://acm.hust.edu.cn:8080/judge/contest/view.action?cid=9948#problem/D

题意：就是给你一种队列，两端都能插入，两端都能删除，然后给你一些操作，输出操作的结果。

思路：用deque可以完美实现，题目没什么难度。。。

代码：

#include <iostream>#include <cstdio>#include <string.h>#include <string>#include <deque>using namespace std;int main(){//freopen("1.txt","r",stdin);int numcase;scanf("%d",&numcase);    for(int K = 1; K <= numcase; ++K){  int numlen,numopera,x;  scanf("%d%d",&numlen,&numopera);  string ss;  deque<int> qq;  printf("Case %d:\n",K);  while(numopera--){    cin>>ss;if(ss == "pushLeft"){scanf("%d",&x);if(qq.size() < numlen){   qq.push_front(x);   printf("Pushed in left: %d\n",x);}else{  printf("The queue is full\n");}}else if(ss == "pushRight"){  scanf("%d",&x);  if(qq.size() < numlen){  qq.push_back(x);  printf("Pushed in right: %d\n",x);  }  else{    printf("The queue is full\n");  }}else if( ss == "popLeft" ){if(qq.size() > 0){  x = qq.front();  qq.pop_front();  printf("Popped from left: %d\n",x);}else{  printf("The queue is empty\n");}}else if( ss == "popRight"){if(qq.size() > 0){x = qq.back();qq.pop_back();printf("Popped from right: %d\n",x);}else{  printf("The queue is empty\n");}}  }}return 0;}