mutiset HDOJ 5349 MZL's simple problem
/*
这题可以用stl的mutiset容器方便求解,我对这东西不熟悉,TLE了几次,最后用读入外挂水过。
题解有O(n)的做法,还以为我是侥幸过的,后来才知道iterator it写在循环内才超时了,囧!
*/
/************************************************
Author :Running_Time
Created Time :2015-8-4 12:10:11
File Name :G.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + ;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + ;
multiset<int>::iterator it; inline int read(void)
{
int x = , f = ; char ch = getchar ();
while (ch < '' || ch > '') {if (ch == '-') f = -; ch = getchar ();}
while (ch >= '' && ch <= '') {x = x * + ch - ''; ch = getchar ();}
return x * f;
} int main(void) { //HDOJ 5349 MZL's simple problem
int n;
while (scanf ("%d", &n) == ) {
multiset<int> S;
for (int i=; i<=n; ++i) {
int op, x;
op = read ();
if (op == ) {
x = read ();
S.insert (x);
}
else if (op == ) {
if (S.empty ()) continue;
S.erase (S.begin ());
}
else if (op == ) {
if (S.empty ()) {
puts (""); continue;
}
it = S.end (); it--;
printf ("%d\n", *it);
}
}
} return ;
}
/************************************************
* Author :Running_Time
* Created Time :2015-8-5 16:42:43
* File Name :G_2.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + ;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + ; inline int read(void)
{
int x = , f = ; char ch = getchar ();
while (ch < '' || ch > '') {if (ch == '-') f = -; ch = getchar ();}
while (ch >= '' && ch <= '') {x = x * + ch - ''; ch = getchar ();}
return x * f;
} int main(void) {
int n; n = read ();
int sz = , mx = -2e9;
for (int i=; i<=n; ++i) {
int op, x; op = read ();
if (op == ) {
x = read ();
sz++; mx = max (mx, x);
}
else if (op == ) {
sz = max (, sz - );
if (!sz) mx = -2e9;
}
else {
if (!sz) puts ("");
else printf ("%d\n", mx);
}
} return ;
}
标程做法
mutiset HDOJ 5349 MZL's simple problem的更多相关文章
- hdoj 5349 MZL's simple problem
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5349 #include<stdio.h> int main(){ int cnt; int ...
- hdu 5349 MZL's simple problem
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5349 MZL's simple problem Description A simple proble ...
- 2015 Multi-University Training Contest 5 hdu 5349 MZL's simple problem
MZL's simple problem Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Oth ...
- 【HDOJ】4267 A Simple Problem with Integers
树状数组.Easy. /* 4267 */ #include <iostream> #include <string> #include <map> #includ ...
- POJ 3468 A Simple Problem with Integers(线段树 成段增减+区间求和)
A Simple Problem with Integers [题目链接]A Simple Problem with Integers [题目类型]线段树 成段增减+区间求和 &题解: 线段树 ...
- POJ 3468 A Simple Problem with Integers(线段树/区间更新)
题目链接: 传送门 A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Description Yo ...
- poj 3468:A Simple Problem with Integers(线段树,区间修改求和)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 58269 ...
- ACM: A Simple Problem with Integers 解题报告-线段树
A Simple Problem with Integers Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%lld & %l ...
- poj3468 A Simple Problem with Integers (线段树区间最大值)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 92127 ...
随机推荐
- 新装mvn建第一个项目报错org.apache.maven.plugins:maven-resources-plugin:2.6
1.第一次创建mvn项目会报maven-resources-plugin-2.6.jar错,原因是mvn无法自动下载这个jar包,多次删除这个目录下的C:\Users\Administrator\.m ...
- Linux下汇编语言学习笔记12 ---
这是17年暑假学习Linux汇编语言的笔记记录,参考书目为清华大学出版社 Jeff Duntemann著 梁晓辉译<汇编语言基于Linux环境>的书,喜欢看原版书的同学可以看<Ass ...
- Charm Bracelet-POJ3624(01背包)
http://poj.org/problem?id=3624 Charm Bracelet Time Limit: 1000MS Memory Limit: 65536K Total Submis ...
- MongoDB学习day03--索引和explain分析查询速度
一.索引基础 db.user.ensureIndex({"username":1}) 创建索引,username为key,数字 1 表示 username 键的索引按升序存储, - ...
- 【CV论文阅读】An elegant solution for subspace learning
Pre: It is MY first time to see quite elegant a solution to seek a subspace for a group of local fea ...
- 牛腩新闻系统(一)——UML、数据库设计
牛腩新闻系统(一)--UML.数据库设计 一.初识牛腩系统 牛腩(Brisket)即牛腹部及靠近牛肋处的松软肌肉,是指带有筋.肉.油花的肉 块.这是一种统称. 若依部位来分,牛身上很多地方的肉都能够叫 ...
- python 区块链程序
python 区块链程序 学习了:https://mp.weixin.qq.com/s?__biz=MzAxODcyNjEzNQ==&mid=2247484921&idx=1& ...
- 【LeetCode-面试算法经典-Java实现】【067-Add Binary(二进制加法)】
[067-Add Binary(二进制加法)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given two binary strings, return thei ...
- MyEclipse 8.5安装Aptana
Aptana简单介绍 Aptana是一个很强大,开源,专注于JavaScript的Ajax开发IDE它的特性包含: 1.JavaScript,JavaScript函数,HTML,CSS语言的Co ...
- FFmpeg的HEVC解码器源码简单分析:概述
===================================================== HEVC源码分析文章列表: [解码 -libavcodec HEVC 解码器] FFmpeg ...