Two Shuffled Sequences

C. Two Shuffled Sequences

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Two integer sequences existed initially — one of them was strictly increasing, and the other one — strictly decreasing.

Strictly increasing sequence is a sequence of integers [x1<x2<⋯<xk]. And strictly decreasing sequence is a sequence of integers [y1>y2>⋯>yl]. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

They were merged into one sequence a. After that sequence a got shuffled. For example, some of the possible resulting sequences a for an increasing sequence [1,3,4] and a decreasing sequence [10,4,2] are sequences [1,2,3,4,4,10] or [4,2,1,10,4,3].

This shuffled sequence a is given in the input.

Your task is to find any two suitable initial sequences. One of them should be strictly increasing and the other one — strictly decreasing. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

If there is a contradiction in the input and it is impossible to split the given sequence a to increasing and decreasing sequences, print “NO”.

Input
The first line of the input contains one integer n (1≤n≤2⋅105) — the number of elements in a.

The second line of the input contains n integers a1,a2,…,an (0≤ai≤2⋅105), where ai is the i-th element of a.

Output
If there is a contradiction in the input and it is impossible to split the given sequence a to increasing and decreasing sequences, print “NO” in the first line.

Otherwise print “YES” in the first line and any two suitable sequences. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

In the second line print ni — the number of elements in the strictly increasing sequence. ni can be zero, in this case the increasing sequence is empty.

In the third line print ni integers inc1,inc2,…,incni in the increasing order of its values (inc1<inc2<⋯<incni) — the strictly increasing sequence itself. You can keep this line empty if ni=0 (or just print the empty line).

In the fourth line print nd — the number of elements in the strictly decreasing sequence. nd can be zero, in this case the decreasing sequence is empty.

In the fifth line print nd integers dec1,dec2,…,decnd in the decreasing order of its values (dec1>dec2>⋯>decnd) — the strictly decreasing sequence itself. You can keep this line empty if nd=0 (or just print the empty line).

ni+nd should be equal to n and the union of printed sequences should be a permutation of the given sequence (in case of “YES” answer).

Examples
inputCopy
7
7 2 7 3 3 1 4
outputCopy
YES
2
3 7
5
7 4 3 2 1
inputCopy
5
4 3 1 5 3
outputCopy
YES
1
3
4
5 4 3 1
inputCopy
5
1 1 2 1 2
outputCopy
NO
inputCopy
5
0 1 2 3 4
outputCopy
YES
0

5
4 3 2 1 0

思路：两个数组，一个递增，一个递减，随机合成一起，现在给你这个数组，问是否可以拆出去

因为这两个数组完全递增或递减，所以合成的数组中不可能出现三个相同的数

我这个写的很多但是应该挺好懂

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int a[200005];//用来存合成序列
int b[200005];//用来存递增序列
int c[200005];//用来存递减序列
int main(){int n;int i;while(scanf("%d",&n)!=EOF){int count1=0;//用来存递减序列的个数int count2=0;//用来存递增序列的个数int flag=0;//用来判断数组中是否有三个相同的数int w=0;//数组c的下标int v=0;//数组b的下标for(i=0;i<n;i++){scanf("%d",&a[i]);}sort(a,a+n); //sort排序if(n==1){        //如果n==1的时候，就直接去按格式输出printf("YES\n");      //别忘了YES，因为这个我WA了四次！！！！！！，就是它！！！！！printf("0\n");                           //哭唧唧printf("\n");                            //想到了伤心往事，后面的注释不写了，自己看去！printf("1\n");printf("%d\n",a[0]);continue;}for(i=0;i<n-2;i++){if(a[i]==a[i+1]){if(a[i]==a[i+2]){flag=1;break;}}}if(flag==1){printf("NO\n");continue;}else{printf("YES\n");}for(i=0;i<n-1;i++){if(a[i]!=a[i+1]){c[w]=a[i];count1++;w++;if(i==n-2){c[w]=a[i+1];count1++;}}else{c[w]=a[i];b[v]=a[i+1];i++;count1++;count2++;w++;v++;if(i==n-2){c[w]=a[i+1];count1++;}}}if(count2==0){printf("0\n");printf("\n");}else{printf("%d\n",count2);for(i=0;i<count2-1;i++){printf("%d ",b[i]);}printf("%d\n",b[count2-1]);}printf("%d\n",count1);for(i=count1-1;i>0;i--){printf("%d ",c[i]);}printf("%d\n",c[0]);}return 0;
}

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