Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) A（贪心）

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最新推荐文章于 2024-09-30 05:54:51 发布

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分类专栏：贪心算法 Codeforces 文章标签： codeforces

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本文介绍了一个基于排序和双指针技术的算法问题，旨在找出序列中满足特定条件的元素数量。通过一次扫描即可高效解决问题。

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A. Oath of the Night’s Watch
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
“Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night’s Watch, for this night and all the nights to come.” — The Night’s Watch oath.

With that begins the watch of Jon Snow. He is assigned the task to support the stewards.

This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.

Can you find how many stewards will Jon support?

Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.

Second line consists of n space separated integers a1, a2, …, an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.

Output
Output a single integer representing the number of stewards which Jon will feed.

Examples
input
2
1 5
output
0
input
3
1 2 5
output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.

In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.

题意：N代表接下来一行有N个数字，找出满足条件的数字X，条件是给定序列中既有比X小的数也有比X大的数，求出满足条件的X的数量。

思路：给定序列排个序，左右两头指针向中间扫描，直至遇到首个满足条件的数停止。

代码:

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
using namespace std;
const int maxn=100005;
int num[maxn];
int main()
{
    int N;
    scanf("%d",&N);
    for(int i=1;i<=N;i++)
        scanf("%d",&num[i]);
    sort(num+1,num+N+1);
    int left=1,right=N;
    while(left<=right&&num[left+1]==num[left])
        left++;
    while(right>=left&&num[right-1]==num[right])
        right--;
    int result=right-left+1;
    if(result>=3)
        printf("%d\n",result-2);
    else
        printf("0\n");
    return 0;
}