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][x1<x2<⋯<xk]. And strictly decreasing sequence is a sequence of integers [y1>y2>⋯>yl][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 aaaaaa[1,3,4][1,3,4][10,4,2][10,4,2][1,2,3,4,4,10][1,2,3,4,4,10][4,2,1,10,4,3][4,2,1,10,4,3]
This shuffled sequence aa 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 aa to increasing and decreasing sequences, print "NO".
The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in aa.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤2⋅1050≤ai≤2⋅105), where aiai is the ii-th element of aa.
If there is a contradiction in the input and it is impossible to split the given sequence aa
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 nininini
In the third line print niniinc1,inc2,…,incniinc1,inc2,…,incniinc1<inc2<⋯<incniinc1<inc2<⋯<incnini=0ni=0
In the fourth line print ndnd — the number of elements in the strictly decreasing sequence. ndnd can be zero, in this case the decreasing sequence is empty.
In the fifth line print ndnddec1,dec2,…,decnddec1,dec2,…,decnddec1>dec2>⋯>decnddec1>dec2>⋯>decndnd=0nd=0
ni+ndni+ndn
代码：
1 #include<iostream> 2 #include<algorithm> 3 #include<set> 4 using namespace std; 5 int main() { 6 int zen[1000000],b[1000000],c[1000000]; 7 int n,zhi,count=0,jishu2=0; 8 cin>>n; 9 set<int> a,d; 10 for(int i=0; i<n; i++) { 11 cin>>zhi; 12 if(a.count(zhi)) { 13 c[jishu2++]=zhi; 14 if(d.count(zhi)) { 15 cout<<"No"; 16 return 0; 17 } 18 d.insert(zhi); 19 } 20 a.insert(zhi); 21 b[i]=zhi; 22 } 23 for(set<int>::iterator it=a.begin(); it!=a.end(); it++) { 24 zen[count++]=*it; 25 } 26 sort(zen,zen+count,greater<int>()); 27 sort(c,c+jishu2); 28 cout<<"Yes"<<endl; 29 cout<<jishu2<<endl; 30 for(int i=0; i<jishu2; i++) { 31 cout<<c[i]<<" "; 32 } 33 cout<<endl; 34 cout<<count<<endl; 35 for(int i=0; i<count; i++) { 36 cout<<zen[i]<<" "; 37 } 38 cout<<endl; 39 }
思路分析：先将所有数存集合中，然后将集合中数存到个数组中再对数组从大到小排序。然后便利，将集合中已存在元素存到新数组中作为递增序列。