Odd Gnome【枚举】
问题 I: Odd Gnome
时间限制: 1 Sec 内存限制: 128 MB
提交: 234 解决: 144
[提交] [状态] [命题人:admin]
题目描述
According to the legend of Wizardry and Witchcraft, gnomes live in burrows underground, known as gnome holes. There they dig
up and eat the roots of plants, creating little heaps of earth around gardens, causing considerable damage to them.
Mrs. W, very annoyed by the damage, has to regularly de-gnome her garden by throwing the gnomes over the fence. It is a lot of work to throw them one by one because there are so many. Fortunately, the species is so devoted to their kings that each group always follows its king no matter what. In other words, if she were to throw just the king over the fence, all the other gnomes in that group would leave.
So how does Mrs. W identify the king in a group of gnomes? She knows that gnomes travel in a certain order, and the king, being special, is always the only gnome who does not follow that order.
Here are some helpful tips about gnome groups:
• There is exactly one king in a group.
• Except for the king, gnomes arrange themselves in strictly increasing ID order.
• The king is always the only gnome out of that order.
• The king is never the first nor the last in the group, because kings like to hide themselves.
Help Mrs. W by finding all the kings!
输入
The input starts with an integer n, where 1 ≤ n ≤ 100, representing the number of gnome groups. Each of the n following lines contains one group of gnomes, starting with an integer g, where 3 ≤ g ≤ 1 000,representing the number of gnomes in that group. Following on the same line are g space-separated integers, representing the gnome ordering. Within each group all the integers (including the king) are unique and in the range [0, 10 000]. Excluding the king, each integer is exactly one more than the integer preceding it.
输出
For each group, output the king’s position in the group (where the first gnome in line is number one).
样例输入
3 7 1 2 3 4 8 5 6 5 3 4 5 2 6 4 10 20 11 12
样例输出
5 4 2 题意: 给出一个序列,序列中有一个King,除了King其他都是递增的,找出King的位置 King不会在第一个和最后一个的位置 那很明显就是枚举2--n-1的位置,找出满足以下两个条件的位置 1.a[i-1] < a[i+1] 2.a[i] > a[i + 1] && a[i] > a[i - 1] 或者 a[i - 1] < a[i + 1]&& a[i] < a[i + 1]
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;
const int maxn = 1000005;
int a[maxn];
int n;
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = 2; i < n; i++)
if ((a[i] > a[i + 1] && a[i] > a[i - 1] && a[i - 1] < a[i + 1] )
||( a[i - 1] < a[i + 1]&& a[i] < a[i + 1] && a[i] < a[i - 1])) {
printf("%d\n", i);
break;
}
}
return 0;
}
Odd Gnome【枚举】的更多相关文章
- upc组队赛6 Odd Gnome【枚举】
Odd Gnome 题目描述 According to the legend of Wizardry and Witchcraft, gnomes live in burrows undergroun ...
- poj 2010 Moo University - Financial Aid(优先队列(最小堆)+ 贪心 + 枚举)
Description Bessie noted that although humans have many universities they can attend, cows have none ...
- Codeforces Round #417 (Div. 2)A B C E 模拟 枚举 二分 阶梯博弈
A. Sagheer and Crossroads time limit per test 1 second memory limit per test 256 megabytes input sta ...
- CodeForces 558C Amr and Chemistry (位运算,数论,规律,枚举)
Codeforces 558C 题意:给n个数字,对每一个数字能够进行两种操作:num*2与num/2(向下取整),求:让n个数相等最少须要操作多少次. 分析: 计算每一个数的二进制公共前缀. 枚举法 ...
- Avito Cool Challenge 2018 E. Missing Numbers 【枚举】
传送门:http://codeforces.com/contest/1081/problem/E E. Missing Numbers time limit per test 2 seconds me ...
- 【枚举+贪心】POJ2718-Smallest Difference
[题目大意] 按升序输出几个不同的数字,任意组成两个数字,输出最小的差值. [思路] 虽然是在穷竭搜索的章节里找到的题目,但是我觉得不需要穷竭搜索,枚举一下就可以了,0MS.分为一下三种情况: (1) ...
- Spoj-ODDDIV Odd Numbers of Divisors
Given a positive odd integer K and two positive integers low and high, determine how many integers b ...
- ARC116 A Odd vs Even (质因数分解,结论)
题面 有 T T T 组数据,每次给出一个数 N N N ,问 N N N 的所有因数(包括 1 1 1 和 N N N)中奇因数个数和偶因数个数的关系(">"," ...
- Swift enum(枚举)使用范例
//: Playground - noun: a place where people can play import UIKit var str = "Hello, playground& ...
随机推荐
- 【转载】解决gridview空行时不显示的问题
问题: GridView控件应用很是广泛,通常将它与DataSourceControl搭配使用,当然也可以手工指定DataSource属性来完成数据绑定.如果数据源返回一个空行的数据集(例如查询不到指 ...
- HTML、CSS知识点,面试开发都会需要--No.5 文章段落
No.5 文章段落 1.文字属性 文字属性包含font-*和text-*两类. 2.基于font的属性 (1)font-family:字体属性,多个字体之前用逗号隔开.如果第一个字体没找到,则依次找后 ...
- layer[申明:来源于网络]
layer[申明:来源于网络] 地址:http://layer.layui.com/
- PAT甲级1135 Is It A Red-Black Tree?【dfs】
题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805346063728640 题意: 给定一棵二叉搜索树的先序遍历结 ...
- Codeforces 1089E - Easy Chess - [DFS+特判][2018-2019 ICPC, NEERC, Northern Eurasia Finals Problem E]
题目链接:https://codeforces.com/contest/1089/problem/E Elma is learning chess figures. She learned that ...
- CentOS 7.6 安装 Weblogic 12
http://download.oracle.com/otn/nt/middleware/12c/12213/fmw_12.2.1.3.0_wls_Disk1_1of1.zip java -jar f ...
- apache tomcat (catalina)查版本(solaris/unix)
先进到tomcat的bin目录下(cd /tomcat目录/bin),在执行./version.sh https://blog.csdn.net/vv___/article/details/78653 ...
- Spring Boot 配置_yaml语法介绍 day02
一.Spring Boot 的全局配置文件(application.properties.application.yml) 1.这两种配置文件是SpringBoot 支持的对默认配置修改的格式.命名和 ...
- Qt中信号(signal)和槽(slot)的几种关联方法
声明,个人总结,不一定正确! 1.最常见的,使用connect语句.比如:connect(btnSend,SIGNAL(clicked()),this,SLOT(clear()); 2.在 .ui设计 ...
- python 模块大全
logging time datetime sys os json random hashlib paramiko pymysql模块使用 subprocess pywi ...