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
- 我永远无法学会的dp
起源:在codeforceround518之后我发现别人都会div1A我根本写不出来,所以我决定退役 咕咕咕咕
- mui 事件管理及自定义事件详解
<!DOCTYPE html> <html> <head> <meta charset="utf-8"> <meta name ...
- myEclipse导入现成项目出现错误 【申明来源于网络】
myEclipse导入现成项目出现错误 [申明来源于网络] 原地址:http://blog.sina.com.cn/s/blog_6d7703400100znh6.html file–>impo ...
- liunx系统问题总结
1.Unable to locate package错误 解决办法 :输入命令 sudo apt-get update,进行软件的更新
- Linux:使用rpcgen实现64位程序调用32位库函数
摘要:本文介绍使用rpcgent实现64位程序调用32位库函数的方法,并给出样例代码. 我的问题 我的程序运行在64位Linux系统上,需要使用一个从外部获得的共享库中的函数,这个共享库是32位的,无 ...
- 《Maven实战》文字版[PDF]
从亚马逊买的电子书,导出来的,需要的下吧. 下面是截图: 除了代码部分有一点点不清楚之外,其他还是蛮清楚的. 下载地址: http://download.csdn.net/download/apple ...
- UI自动化框架——构建思维
目的:从Excel中获取列的值,传输到页面 技巧:尽可能的提高方法的重用率 Java包: 1.java.core包 3个类:1)日志(LogEventListener)扩展web driver自带的事 ...
- QSS独门秘籍:subcontrol
QSS是C++ Qt中的界面美化神器,其语法和CSS区别不大,但是QSS有一个独有的功能——subcontrol,这是CSS所没有的,一个widget往往由多个子部件构成,利用subcontrol可以 ...
- 配置DNS Server容易忽略的问题
1.named服务启动成功,但nslookup解析报错: [root@xiamihost3 named]# service named restart 停止 named: [确定] 启动 named: ...
- python练习题-day12
用列表推导式做下列小题 (1) 过滤掉长度小于3的字符串列表,并将剩下的转换成大写字母 lst1=["admhdja","aksaudj","fh&q ...