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
- react Context
import React, { useState, useEffect, useContext } from "react"; import axios from "ax ...
- laravel5.4将excel表格中的信息导入到数据库中
本功能是借助 Maatwebsite\Excel 这个扩展包完成的,此扩展包的安装过程请参考上篇博文:http://www.cnblogs.com/zhuchenglin/p/7122946.html ...
- Palindromic Matrix
Palindromic Matrix time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- 用js实现二维数组的旋转
我最近因为做了几个小游戏,用到了二维数组,其中有需求将这个二维数组正翻转 90°,-90°,180°. 本人是笨人,写下了存起来. 定义的基本二位数组渲染出来是这种效果. 现在想实现的结果是下面的效果 ...
- Winscp 密钥登录
继前段时间,使用密钥SSH登录系统以后运维用着很爽,但是有一部分开发的同事反应使用Winscp添加私钥的时候,提示添加失败,原来Winscp使用的是putty作为SSH登录工具,而puttygen所生 ...
- GetAsyncKeyState()& 0x8000
0x8000 & GetKeyState(VK_SHIFT); 这句是判断是否有按下shift键. 关于GetAsyncKeyState与GetKeyState区别:关于GetAsyncKey ...
- (转)JVM中的OopMap(zz)
原文地址: http://www.cnblogs.com/strinkbug/p/6376525.html 在读周智明的深入理解JVM虚拟机时,关于枚举根节点/安全点这部分感觉书上写的不是太明白,找了 ...
- WordCount扩展
码云地址:https://gitee.com/xjtsh/ExpandedWordCount 功能实现: wc.exe -c file.c //返回文件 file.c 的字符数 wc.exe ...
- 队列模拟基本操作I
看到这道题,第一个想法就是“搜索”!“回溯”!的确,这种思路是很正确的,BFS和DFS都可以来解决: #include <cstdlib> #include <cstring> ...
- linux 系统下apache 找不到apxs 文件
yum install httpd-devel