Two progressions CodeForce 125D 思维题

Posted 2022-12-12 cautx

An arithmetic progression is such a non-empty sequence of numbers where the difference between any two successive numbers is constant. This constant number is called common difference. For example, the sequence 3, 7, 11, 15 is an arithmetic progression. The definition implies that any sequences whose length equals 1 or 2 are arithmetic and all sequences whose length equals 0 are non-arithmetic.

You are given a sequence of different integers a₁, a₂, ..., a_n. You should either split it into two arithmetic progressions or find out that the operation is impossible to perform. Splitting assigns each member of the given sequence to one of two progressions, but the relative order of numbers does not change. Splitting is an inverse operation to merging.

Input

The first line contains a positive integer n (2 ≤ n ≤ 30000), n is the length of the given sequence. The second line contains elements of the given sequence a₁, a₂, ..., a_n ( - 10⁸ ≤ a_i ≤ 10⁸). The elements of the progression are different integers.

Output

Print the required arithmetic progressions, one per line. The progressions can be positioned in any order. Each progression should contain at least one number. If there‘s no solution, then print "No solution" (without the quotes)in the only line of the input file. If there are several solutions, print any of them.

Examples

Input

6
4 1 2 7 3 10

Output

1 2 3 
4 7 10 

Input

5
1 2 3 -2 -7

Output

1 2 3 
-2 -7 

Note

In the second sample another solution is also possible (number three can be assigned to the second progression): 1, 2 and 3, -2, -7.

OJ-ID：
CodeForce 125D

author：
Caution_X

date of submission：
20191002

tags：
思维

description modelling：
给定一个序列，把它分成两个非空子序列s1,s2,(s1+s2=全集)，每一个序列都是等差序列，如果可以输出两个子序列，否则输出No solution

解：
(1) 对每个数，要么在第一个子序列，要么在第二个子序列，根据鸽巢原理，将前三个元素放入两个数列必然有两个数在同一个数列，其中元素个数大于1的数列会形成一个公差，且公差d最多只有有三个值
(2) 根据其中一个公差d生成一个等差数列，然后将剩下的数放在另一个数列，判断另一个数列是不是等差数列
(3) 是等差数列则直接输出，否则将第一个数列尾部元素移到第二个数列，判断是否构成等差数列
(4) 如果构成则输出，否则枚举其他公差值

补充：
对(3)->(4)中将其中一个数列尾部元素移至另一个数列时若还不构成等差数列则应该枚举其他公差的证明：
从构造的等差数列里移动2个元素过来，假设，移动第一个后没有形成等差数列，移动第二个后形成了等差数列，那么这个新形成的等差数列公差和原来构建的等差数列的公差相等，也就是说这个新形成的等差数列和原来的等差数列可以放在同一个数列中，即：在拆成两个子序列之前，原数列就是等差数列,既然原数列是等差数列，那么在移动第一元素后新数列就已经是等差数列了，和我们假设的移动第一个后没有形成等差数列矛盾，因此得出：第一元素移动后没有形成等差数列，那么接下来无论移动多少个元素都不会形成等差数列，因此当第一个元素移动完之后若不是等差数列就直接枚举其他的公差。

 

AC CODE:

#include<bits/stdc++.h>
using namespace std;
int a[30005],n;
bool vis[30005];
void print(vector<int> v)

    for(int i=0; i<v.size(); i++)    printf("%d ",v[i]);
    printf("\n");

bool check(vector<int> v)

    if(v.empty())    return false;
    else if(v.size()==1||v.size()==2)    return true;
    int d=v[1]-v[0];
    for(int i=1; i<v.size(); i++) 
        if(v[i]-v[i-1]!=d)    return false;
    
    return true;

bool solve(int l,int r)

    vector<int> v1,v2;
    int d=a[r]-a[l],get=a[l],last=-1;
    for(int i=0; i<=n; i++)    vis[i]=false;
    for(int i=1; i<=n; i++) 
        if(a[i]==get) 
            get+=d;
            v1.push_back(a[i]);
            last=i;
         else 
            vis[i]=true;
        
    
    for(int i=1; i<=n; i++) 
        if(vis[i]) 
            v2.push_back(a[i]);
        
    
    if(check(v2)) 
        print(v1);
        print(v2);
        return true;
     else 
        v1.pop_back();
        v2.clear();
        vis[last]=true;
        for(int i=1; i<=n; i++) 
            if(vis[i])
                v2.push_back(a[i]);
        
        if(check(v2)) 
            print(v1);
            print(v2);
            return true;
        
    
    return false;

int main()

    //freopen("input1.txt","r",stdin);
    //freopen("input2.txt","r",stdin);
    scanf("%d",&n);
    for(int i=1; i<=n; i++)    
        scanf("%d",&a[i]);
    
    if(n==2)    printf("%d\n%d",a[1],a[2]);
    else if(!solve(1,2)&&!solve(1,3)&&!solve(2,3))    printf("No solution\n");
    return 0;