[简单思维题]Sequence(山东省第九届ACM大学生程序设计竞赛E题)

Problem Description

We define an element $a_iai in a sequence "good", if and only if there exists a j(1\le j < i)j(1≤j<i) such that a_j < a_iaj<ai.Given a permutation pp of integers from 11 to nn. Remove an element from the permutation such that the number of "good" elements is maximized.$

Input

The input consists of several test cases. The first line of the input gives the number of test cases, $T(1\le T\le 10^3)T(1≤T≤103).For each test case, the first line contains an integer n(1\le n\le 10^6)n(1≤n≤106), representing the length of the given permutation.The second line contains nn integers p_1,p_2,\cdots,p_n(1\le p_i\le n)p1,p2,⋯,pn(1≤pi≤n), representing the given permutation pp.It’s guaranteed that \sum n\le 2\times 10^7∑n≤2×107.$

Output

For each test case, output one integer in a single line, representing the element that should be deleted. If there are several answers, output the minimal one.

Sample Input

Sample Output

1
5

思路：首先要明白，一个不好数，它总是它之前(包括它)出现的所有数中最小的；
考虑删的删除一个好数还是不好数：
删除一个好数，则总的好数数量将减少一个(因为删除它后能影响到的好数仅有它自己)；
删除一个不好数，考虑删除这个不好数后能影响到的好数有几个，那么总的好数数量就减少几个；(被它所影响到的好数(假设目前考虑的好数是a[i])是那些满足 最小{a[0~1-i]}<a[i]<=次小{a[0~n-1]} 的数)

AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
typedef long long ll;
#define inf 0x3f3f3f3f
using namespace std;

ll a[1000010];
ll first[1000010];
ll second[1000010];
bool flag[1000010];
ll cnt[1000010];

int main()
{
    ll t;
    scanf("%lld",&t);
    while(t--){
        ll n;
        scanf("%lld",&n);
        ll now_min=inf;ll pre_min=inf;
        for(ll i=1;i<=n;i++){
            scanf("%lld",&a[i]);
            first[i]=now_min; second[i]=pre_min;
            //维护前缀最小和前缀次小
            if(a[i]<=now_min) {pre_min=now_min; now_min=a[i];}
            else {if(a[i]<pre_min) pre_min=a[i];}
        }
        for(ll i=1;i<=n;i++){//不好数标记为1，好数标记为0
            if(a[i]<=first[i]) flag[i]=1;
            else flag[i]=0;
        }
        ll k;
        for(ll i=1;i<=n;i++){
            if(flag[i]==1){
                k=i;
                cnt[k]=0;
            }
            else{
                if(a[i]<=second[i]) cnt[k]++;
            }
        }
        ll ans=inf;
        for(ll i=1;i<=n;i++){
            if(flag[i]==1&&cnt[i]==0) ans=min(ans,a[i]);
        }
        if(ans==inf) {
            for(ll i=1;i<=n;i++){
              if(flag[i]==0) ans=min(ans,a[i]);
              else if(cnt[i]==1) ans=min(ans,a[i]);
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}

过了此题是真的开心~~~，要多思考啊