PAT甲级1057. Stack
PAT甲级1057. Stack
题意:
堆栈是最基础的数据结构之一,它基于“先进先出”(LIFO)的原理。基本操作包括Push(将元素插入顶部位置)和Pop(删除顶部元素)。现在你应该实现一个额外的操作堆栈:PeekMedian -
返回堆栈中所有元素的中间值。对于N个元素,如果N是偶数,则将中值定义为(N / 2)个最小元素,或者如果N是奇数则将其定义为((N + 1)/ 2)。
输入规格:
每个输入文件包含一个测试用例。对于每种情况,第一行包含正整数N(<= 105)。然后N行跟随,
每个都包含以下3种格式之一的命令:
按键
流行的
PeekMedian
其中key是正整数,不超过105。
输出规格:
对于每个Push命令,将密钥插入堆栈并输出任何内容。对于每个Pop或PeekMedian命令,在一行中打印相应的返回值。如果命令无效,
打印“无效”。
思路:
树状数组 参考于:
PAT1057.Stack (30)
树状数组
ac代码:
C++
// pat1057.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstring>
#include<stdio.h>
#include<map>
#include<cmath>
#include<unordered_map>
using namespace std;
//树形数组
const int maxn = 1e5 + 5;
int c[maxn];
int lowbit(int i) //count 2^k, (k 为数字二进制末尾0的个数)
{
return i & (-i);
}
void add(int pos, int value)
{
while (pos < maxn)
{
c[pos] += value;
pos += lowbit(pos);
}
}
int sum(int pos)
{
int res = 0;
while (pos > 0)
{
res += c[pos];
pos -= lowbit(pos);
}
return res;
}
int find(int value)
{
int l = 0, r = maxn - 1, median, res;
while (l < r - 1)
{
if ((l + r) % 2 == 0)
{
median = (l + r) / 2;
}
else
{
median = (l + r - 1) / 2;
}
res = sum(median);
if (res < value)
{
l = median;
}
else
{
r = median;
}
}
return l + 1;
}
int main()
{
int n;
scanf("%d", &n);
int stack[maxn], top = 0, pos;
memset(c, 0, sizeof(c));
char oper[20];
while(n--)
{
scanf("%s", oper);
if (oper[1] == 'o') //pop
{
if (top == 0)
{
printf("Invalid\n");
continue;
}
int out = stack[top];
add(out, -1);
printf("%d\n", stack[top--]);
}
else if (oper[1] == 'e') //peekmedian
{
if (top == 0)
{
printf("Invalid\n");
continue;
}
int res;
if (top % 2 == 0) res = find(top / 2);
else res = find((top + 1) / 2);
printf("%d\n", res);
}
else if (oper[1] == 'u') //push
{
scanf("%d", &pos);
stack[++top] = pos;
add(pos, 1);
}
else
{
printf("Invalid\n");
}
}
return 0;
}
PAT甲级1057. Stack的更多相关文章
- PAT 甲级1057 Stack (30 分)(不会,树状数组+二分)*****
1057 Stack (30 分) Stack is one of the most fundamental data structures, which is based on the prin ...
- pat 甲级 1057 Stack(30) (树状数组+二分)
1057 Stack (30 分) Stack is one of the most fundamental data structures, which is based on the princi ...
- PAT甲级1057 Stack【树状数组】【二分】
题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805417945710592 题意:对一个栈进行push, pop和 ...
- PAT 甲级 1057 Stack
https://pintia.cn/problem-sets/994805342720868352/problems/994805417945710592 Stack is one of the mo ...
- PAT甲级——A1057 Stack
Stack is one of the most fundamental data structures, which is based on the principle of Last In Fir ...
- PAT 1057 Stack [难][树状数组]
1057 Stack (30)(30 分) Stack is one of the most fundamental data structures, which is based on the pr ...
- pat甲级题解(更新到1013)
1001. A+B Format (20) 注意负数,没别的了. 用scanf来补 前导0 和 前导的空格 很方便. #include <iostream> #include <cs ...
- PAT甲级题解(慢慢刷中)
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给 ...
- 【转载】【PAT】PAT甲级题型分类整理
最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel P ...
随机推荐
- PHP提取url
<?php $str = parse_url('http://localhost/?id=2&cd=2', PHP_URL_QUERY); ECHO $str; parse_str($s ...
- C++ 内联函数inline
http://blog.csdn.net/u011327981/article/details/50601800 1. 内联函数 在C++中我们通常定义以下函数来求两个整数的最大值: 复制代码 代码 ...
- 94.Binary Tree Inorder Traversal---二叉树中序非递归遍历
题目链接 题目大意:中序遍历二叉树.先序见144,后序见145. 法一:DFS,没啥说的,就是模板DFS.代码如下(耗时1ms): public List<Integer> inorder ...
- centos 下单独安装mysql
https://www.cnblogs.com/running-mydream/p/4666094.html https://www.cnblogs.com/lzj0218/p/5724446.htm ...
- leetcode 之Search in Rotated Sorted Array(四)
描述 Follow up for ”Search in Rotated Sorted Array”: What if duplicates are allowed? Would this aff ...
- vue中methods、watch、computed之间的差别对比以及适用场景
首先要说,methods,watch和computed都是以函数为基础的,但各自却都不同: 一.computer 当页面中有某些数据依赖其他数据进行变动的时候,可以使用计算属性. <p id=& ...
- Diffie–Hellman key exchange
General overview[edit] Illustration of the idea behind Diffie–Hellman key exchange Diffie–Hellman ...
- BNUOJ 52505 Euclidean Geometry
结论. 算了好久不会算,最后看了样例猜出了结论.次长边全用上,再用最长边减去次长边. #include<bits/stdc++.h> using namespace std; int T; ...
- Python并发编程-多进程进程锁
from multiprocessing import Process import json import time from multiprocessing import Lock def sho ...
- 去除win7桌面图标小箭头.bat
reg add "HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Explorer\Shell Icons" ...