codeforces 708 (A-C)

A. Meximization

#include<bits/stdc++.h>
using namespace std;
int a[105]={0};
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		memset(a,0,sizeof(a));
		int n;
		cin>>n;
		for(int i=0;i<n;i++)
		{
			int x;
			cin>>x;
			a[x]++;
		}
		for(int i=0;i<105;i++)
		{
			if(a[i]!=0)
			{
				cout<<i<<" ";
				a[i]--;	
			}
		}
		for(int i=0;i<105;i++)
		{
			while(a[i]!=0)
			{
				cout<<i<<" ";
				a[i]--;	
			}
		}
		cout<<endl;
	}
 } 

B. M-arrays

• 将互余的项放在一起，数量多的比少的最多多一个

#include<bits/stdc++.h>
using namespace std;
const int N = 100005;
int a[N];
int visit[N];
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		memset(a,0,sizeof(a));
		memset(visit,0,sizeof(visit));
		int n,m;
		cin>>n>>m;
		for(int i=0;i<n;i++)
		{
			int x;
			cin>>x;
			a[x%m]++;
		}
		int ans=0;
		for(int i=0;i<m;i++)
		{
			if(a[i]&&!visit[i])
			{
				if(i%m==0)
				{
					ans++;
				}
				else if(!a[m-(i)%m])
				{
					ans+=a[i];	
				}
				else
				{
					int mi=min(a[i],a[m-(i)%m]);
					int ma=max(a[i],a[m-(i)%m]);
					if(mi!=ma)
					{
						ans+=ma-mi;
					}
					else
					ans++;
				}
				visit[m-(i)%m]=1;
			}
		}
		cout<<ans<<endl;
	}
}

C1. k-LCM (easy version)

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n,k;
		cin>>n>>k;
		if(n&1)
		{
			cout<<1<<" "<<(n-1)/2<<" "<<(n-1)/2<<endl;
		}
		else
		{
			if(n%4==2)
			{
				cout<<2<<" "<<(n-2)/2<<" "<<(n-2)/2<<endl;
			}
			else
			{
				cout<<n/4<<" "<<n/4<<" "<<n/2<<endl;
			}
		}
	}
}

C2. k-LCM (hard version)

前面的全部取1，就和easy version一样了

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n,k;
		cin>>n>>k;
		while(k>3)
		{
			n--;
			k--;
			cout<<1<<" ";
		}
		if(n&1)
		{
			cout<<1<<" "<<(n-1)/2<<" "<<(n-1)/2<<endl;
		}
		else
		{
			if(n%4==2)
			{
				cout<<2<<" "<<(n-2)/2<<" "<<(n-2)/2<<endl;
			}
			else
			{
				cout<<n/4<<" "<<n/4<<" "<<n/2<<endl;
			}
		}
	}
}