Codeforces Round #586 (Div. 1 + Div. 2)
D. Alex and Julian
Description
Boy Dima gave Julian a birthday present - set BB consisting of positive integers. However, he didn't know, that Julian hates sets, but enjoys bipartite graphs more than anything else!
Julian was almost upset, but her friend Alex said, that he can build an undirected graph using this set in such way: let all integer numbers be vertices, then connect any two ii and jj with an edge if |i−j||i−j| belongs to BB.
Unfortunately, Julian doesn't like the graph, that was built using BB. Alex decided to rectify the situation, so he wants to erase some numbers form BB, so that graph built using the new set is bipartite. The difficulty of this task is that the graph, Alex has to work with, has an infinite number of vertices and edges! It is impossible to solve this task alone, so Alex asks you for help. Write a program that erases a subset of minimum size from BB so that graph constructed on the new set is bipartite.
Recall, that graph is bipartite if all its vertices can be divided into two disjoint sets such that every edge connects a vertex from different sets.
Input
First line contains an integer n (1⩽n⩽200000)n (1⩽n⩽200000) — size of BB
Second line contains nn integers b1,b2,…,bn (1⩽bi⩽1018)b1,b2,…,bn (1⩽bi⩽1018) — numbers of BB, all bibi are unique
output
In first line print single integer kk – number of erased elements. In second line print kk integers – values of erased elements.
If there are multiple answers, print any of them.
Examples
Input
3
1 2 3
Output
1
2
正确解法：
题目说 如果集合b中有 |i-j| 这个数时，就把 i 和 j 连一条边。
图有无数个点无数个边，求删除尽量少的数使图是一个二分图。
通过画图，发现 当有 2 8 的时候，2和8肯定不能在一起。
这就引入了一个2的k次幂的问题。
当两个数同时除 2^k 时变成了两个奇数。 这就可以了。
二分图必须是偶环。
但是这么多组合怎么知道删除哪个最优呢？
我看到了这个：
1 for(int i=1;i<=n;i++) 2 { 3 scanf("%lld",&num[i]); 4 int k=0; 5 x=num[i]; 6 while(x%2==0) 7 { 8 x/=2; 9 k++; 10 } 11 a[k].push_back(i); 12 }
于是找到最大的那个集合就可以了。
1 #include<iostream> 2 #include<cstdio> 3 #include<string> 4 #include<algorithm> 5 #include<vector> 6 #include<cstring> 7 #include<numeric> 8 #include <cmath> 9 #include<map> 10 using namespace std; 11 typedef long long ll; 12 const int SIZE = 1010; 13 const int mod=1e9+7; 14 const int N=2e5+100; 15 int n,m; 16 ll num[N],x; 17 vector<ll>a[70]; 18 int main() 19 { 20 scanf("%d",&n); 21 for(int i=1;i<=n;i++) 22 { 23 scanf("%lld",&num[i]); 24 int k=0; 25 x=num[i]; 26 while(x%2==0) 27 { 28 x/=2; 29 k++; 30 } 31 a[k].push_back(i); 32 } 33 m=-1; 34 for(int i=0;i<=64;i++) 35 { 36 if(m==-1||a[m].size()<a[i].size()) 37 m=i; 38 } 39 printf("%d ",n-a[m].size()); 40 for(int i=0;i<=64;i++) 41 { 42 if(i==m) continue; 43 for(int j=0;j<a[i].size();j++) 44 printf("%lld ",num[a[i][j]]); 45 } 46 47 48 49 return 0; 50 }