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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 【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) ...

  7. Spoj-ODDDIV Odd Numbers of Divisors

    Given a positive odd integer K and two positive integers low and high, determine how many integers b ...

  8. 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(">"," ...

  9. 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

  1. VUE组件的学习

    参考:https://blog.csdn.net/baidu_23142899/article/details/79130225

  2. js中 函数声明/函数表达式/匿名函数/箭头函数/立即执行函数

    函数声明: function add(a, b) { // ... } 1.顾名思义,声明一个函数, 用关键字 “function” 来告诉,这是一个函数. 2.任何地方,想用就可以拿过来使用 函数表 ...

  3. hibernate07--关联映射

    单向的一对多关联 创建对应的实体类以及映射文件 package cn.bdqn.bean; /** * * @author 小豆腐 *街道对应的实体类 * *单向的多对一关联 */ public cl ...

  4. 24.redux

    Flux:Flux 是一种架构思想 https://facebook.github.io/flux/ 官网 资料: http://www.ruanyifeng.com/blog/2016/01/flu ...

  5. Listen error 错误和 limit of inotify watches was reached

    今天在生产环境中报错rails c中报了一个错误: FATAL: Listen error: unable to monitor directories for changes. Visit http ...

  6. mysql数据库数据的 备份以及还原

    数据库备份的3种方式: 例如:mysqldump -uzx_root -p test>/root/test1.sql

  7. 下载频道--IT资源关东煮第二期[申明:来源于网络]

    下载频道–IT资源关东煮第二期[申明:来源于网络] 地址:http://geek.csdn.net/news/detail/129509?ref=myread

  8. RFID系统 免费开源代码 开发,分享[申明:来源于网络]

    RFID系统 免费开源代码 开发,分享[申明:来源于网络] 地址:http://www.codeforge.cn/s/0/RFID%E7%B3%BB%E7%BB%9F

  9. ios学习--iphone 实现下拉菜单

    原文地址:ios学习--iphone 实现下拉菜单作者:sdglyuan00 #import @interface DropDown1 : UIView <</span>UITabl ...

  10. InnoDB中锁的模式

    Ⅰ.总览 S行级共享锁 lock in share mode X行级排它锁 增删改 IS意向共享锁 IX意向排他锁 AI自增锁 Ⅱ.锁之间的兼容性 兼 X IX S IS X × × × × IX × ...