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
- ci框架(codeigniter)Email发送邮件、收件人、附件、Email调试工具
ci框架(codeigniter)Email发送邮件.收件人.附件.Email调试工具 Email 类 CodeIgniter 拥有强大的 Email 类来提供如下的功能: 多 ...
- JDBC的结果集
以下内容引用自http://wiki.jikexueyuan.com/project/jdbc/result-sets.html: SQL语句从数据库查询中获取数据,并将数据返回到结果集中.SELEC ...
- tomcat8.5.20配置https
一.使用cmd下生成证书: d: cd d:/java/jdk/jdk1.8 keytool -v -genkey -alias tomcat -keyalg RSA -keystore D:\jav ...
- mysql: reinit the password
You can reinit the password : 1.stop mysql /etc/init.d/mysql stop 2.start mysql safe : mysqld_safe - ...
- 旧瓶新酒之ngx_lua & fail2ban实现主动诱捕
服务器承担着业务运行及数据存储的重要作用,因此极易成为攻击者的首要目标.如何对业务服务器的安全进行防护,及时找出针对系统的攻击,并阻断攻击,最大程度地降低主机系统安全的风险程度,是企业安全从业人员面临 ...
- UVa 340 Master-Mind Hints(猜数字游戏的提示)
题意 猜数字游戏 统计猜的数字有多少个数字位置正确 有多少个数字在答案中出现可是位置不对 每一个字符仅仅能匹配一次 直接匹配每位数 #include<cstdio> #includ ...
- Spring——IoC
控制反转(Inversion ofControl,英文缩写为IoC)是一种能够解耦的方法,不是什么技术.是一种思想,也是轻量级的Spring框架的核心.控制反转一般分为两种类型.依赖注入(Depend ...
- tf.image.resize_bilinear 图像缩放,双线性插值-图像中心对齐
http://www.cnblogs.com/yssongest/p/5303151.html 双线性插值算法及需要注意事项 input = tf.placeholder(tf.float32, sh ...
- 第五课 Struts的控制器【续】Action类的execute()方法
1.Action类的execute()方法: public ActionForward execute(ActionMapping mapping, ...
- Android 通过USB查看kernel调试信息【转】
本文转载自:http://blog.csdn.net/lindonghai/article/details/51683644 前提:电脑已安装adb并可正常使用. 在调试Android驱动时,需要查看 ...