NewOJ Week 6

B 特殊数字—预处理+二分

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>

using namespace std;
#define int long long
const int N = 100010;
vector<int>nums;
signed main()
{
    
	for(int i=0;i<=30;i++)
		for(int j=i+1;j<=30;j++)
			nums.push_back((1<<i)+(1<<j));
	sort(nums.begin(),nums.end());
	
	int T;cin>>T;
	while(T--)
	{
    
		int x;cin>>x;
		int res=1e9;
		int pos=lower_bound(nums.begin(),nums.end(),x)-nums.begin();
		if(nums[pos]==x)res=0;
		else
		{
    
			res=nums[pos]-x;
			if(pos>0)res=min(res,x-nums[pos-1]);
		}
		cout<<res<<'\n';
	}
	
}

C 质数拼图游戏—BFS

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define int long long
int dx[4]={
    0,1};
int dy[4]={
    1,0};
int vis[21]={
    0,
	0,1,1,0,1,0,1,0,0,0,
	1,0,1,0,0,0,1,0,1,0,
};
queue<string>q;
map<string,int>mp;
void bfs(string s)
{
    
	q.push(s);
	mp[s]=0;
	while(q.size())
	{
    
		string t=q.front();
		q.pop();
		for(int i=0;i<3;i++)for(int j=0;j<3;j++)
		{
    
			int x1=i*3+j;
			for(int k=0;k<2;k++)
			{
    
				int x=i+dx[k];
				int y=j+dy[k];
				if(x>=3||y>=3)continue;
				int x2=x*3+y;
				if(vis[t[x1]-'0'+t[x2]-'0'])
				{
    
					string ss=t;
					swap(ss[x1],ss[x2]);
					if(!mp.count(ss))
					{
    
						mp[ss]=mp[t]+1;
						q.push(ss);
					}
				}
			}
		}
	}
}

signed main()
{
    
	string s="123456789";
	bfs(s);
	int T;cin>>T;
	while(T--)
	{
    
		string ss="";
		for(int i=0;i<9;i++)
		{
    
			char op;cin>>op;
			ss+=op;
		}
		if(mp.count(ss))cout<<mp[ss]<<'\n';
		else cout<<-1<<'\n';
	}
}

D 推箱子—差分

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define int long long
const int N=1000010;
int s[N]={
    0}; 
signed main()
{
    
	int n,h;cin>>n>>h;
	for(int i=1,l,r;i<=n;i++)
	{
    
		cin>>l>>r;
		s[l]++;
		s[r+1]--;
	}
	for(int i=1;i<=n;i++)s[i]+=s[i-1];
	for(int i=1;i<=n;i++)s[i]+=s[i-1];
	int res=1e18;
	for(int i=1;i+h-1<=n;i++)
	{
    
		int t=s[i+h-1]-s[i-1];
		res=min(res,n*h-t);
	}
	cout<<res<<'\n';
}

E 最小公倍数—数学推导

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<cmath>
using namespace std;
#define int long long
const int N=1000010 , INF=1e18;
map<int,pair<int,int>>mp;
void init()
{
    
	for(int l=1;l<=2000000;l++)
	{
    
		int sum=l*(l+1);
		for(int r=l+2;;r++)
		{
    
			int g=__gcd(sum,r);
			if(sum/g>INF/r)break;
			sum=sum/g*r;
			if(!mp.count(sum))
			{
    
				mp[sum]={
    l,r};
			}
		}
	}
}

void solve()
{
    
	pair<int,int>res;
	int n;cin>>n;
	int t=(sqrt(4*n+1)-1)/2;
	if(t*(t+1)==n)
	{
    
		res={
    t,t+1};
		if(mp.count(n))
		{
    
			if(mp[n].first<t)
				res=mp[n];
		}
	}
	else if(mp.count(n))res=mp[n];
	else 
	{
    
		cout<<-1<<'\n';
		return ;
	}
	cout<<res.first<<" "<<res.second<<'\n';
}

signed main()
{
    
	init();
	int T;cin>>T;
	while(T--)solve();
}