Codeforces Round 861 (Div. 2)（A~C）题解

比赛链接：https://codeforces.com/contest/1808

A. Lucky Numbers

思路：这道题如果想去直接构造挺难的，但是我们可以很简单想到0和9在距离一个数的上下1000范围内是一定会出现的，所以就直接枚举即可

#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<string>
#define ll                                long long 
#define lowbit(x) (x & -x)
//#define endl "\n"//                           交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
const ll N = 205000;
#define F first
#define S second
ll ksm(ll x, ll y)
{
	ll ans = 1;
	while (y)
	{
		if (y & 1)
		{
			ans = ans % mod * (x % mod) % mod;
		}
		x = x % mod * (x % mod) % mod;
		y >>= 1;
	}
	return ans % mod % mod;
}
ll gcd(ll x, ll y)
{
	if (y == 0)
		return x;
	else
		return gcd(y, x % y);
}
void fio()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
}
struct s
{
	ll l, r;
	bool operator<(const s& a)
	{
		return r < a.r;
	}
};
int main()
{
	fio();
	ll t;
	cin >> t;
	while (t--)
	{
		ll l,r;
		cin>>l>>r;
		ll pd=0;
		// while(1)
		// {

		// }
		ll ans=-5;
		ll cnt;
		for(ll i=l;i<=min(l+1000,r);i++)
		{
			ll u=i;
			ll minn=0,maxx=1e18;
			while(u)
			{
				minn=max(minn,u%10);
				maxx=min(maxx,u%10);
				u/=10;
			}
			if(ans<abs(maxx-minn))
			cnt=i,ans=abs(maxx-minn);
			//ans=max(ans,abs(maxx-minn));
		}
		for(ll i=r;i>=max(l,r-1000);i--)
		{
			ll u=i;
			ll minn=0,maxx=1e18;
			while(u)
			{
				minn=max(minn,u%10);
				maxx=min(maxx,u%10);
				u/=10;
			}
			if(ans<abs(maxx-minn))
			cnt=i,ans=abs(maxx-minn);
		}
		cout<<cnt<<endl;
	}
}

B. Playing in a Casino

思路：答案是每列操作之和，那我们就对每一列单独计算，可以想到排个序，然后从大往小进行叠加就可以了。

#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<string>
#define ll                                long long 
#define lowbit(x) (x & -x)
//#define endl "\n"//                           交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
const ll N = 205000;
#define F first
#define S second
ll ksm(ll x, ll y)
{
	ll ans = 1;
	while (y)
	{
		if (y & 1)
		{
			ans = ans % mod * (x % mod) % mod;
		}
		x = x % mod * (x % mod) % mod;
		y >>= 1;
	}
	return ans % mod % mod;
}
ll gcd(ll x, ll y)
{
	if (y == 0)
		return x;
	else
		return gcd(y, x % y);
}
void fio()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
}
struct s
{
	ll l, r;
	bool operator<(const s& a)
	{
		return r < a.r;
	}
};
//ll a[450000];
int main()
{
	fio();
	ll t;
	cin >> t;
	while (t--)
	{
		ll n,m;
		cin>>n>>m;
		vector<ll>a[m+5];
		for(ll i=1;i<=n;i++)
		{
			for(ll j=1;j<=m;j++)
			{
				ll x;
				cin>>x;
				a[j].push_back(x);
			}
		}
		for(ll j=1;j<=m;j++)sort(a[j].begin(),a[j].end());
		ll ans=0;
		for(ll j=1;j<=m;j++)
		{
			ll cnt=0;
			ll f=ans;
			ll cs=0;
			for(ll k=(ll)a[j].size()-1;k>=0;k--)
			{
				ll u=a[j][k];
				if(cnt>0)
				ans+=cnt-u*cs;
				cnt+=u;
				cs++;
			}
			//cout<<ans-f<<endl;
		}
		cout<<ans<<endl;

	}
}

C. Unlucky Numbers

思路：可以想到如果左右数字的长度不一样，答案就是999...，不一样，可以dfs暴力，用标记表示是否得被上限或者下限；其实还有更简单的方法，直接找到第一个不想等的区域，然后暴力下这个区间，然后再暴力下后面的区间（后面的默认一样即可）最坏情况,9*9.这种好写多了

#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<string>
#define ll                                long long 
#define lowbit(x) (x & -x)
//#define endl "\n"//                           交互题记得删除
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
const ll N = 205000;
#define F first
#define S second
ll ksm(ll x, ll y)
{
	ll ans = 1;
	while (y)
	{
		if (y & 1)
		{
			ans = ans % mod * (x % mod) % mod;
		}
		x = x % mod * (x % mod) % mod;
		y >>= 1;
	}
	return ans % mod % mod;
}
ll gcd(ll x, ll y)
{
	if (y == 0)
		return x;
	else
		return gcd(y, x % y);
}
void fio()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
}
struct s
{
	ll l, r;
	bool operator<(const s& a)
	{
		return r < a.r;
	}
};
//ll a[450000];
int main()
{
	fio();
	ll t;
	cin >> t;
	while (t--)
	{
		string l, r;
		cin >> l >> r;
		ll len = max(l.size(), r.size());
		if (l.size() != r.size())
		{
			for (ll i = 0; i < l.size(); i++)cout << "9";
			cout << endl;
			continue;
		}
		ll fg[250], an, e = 1e18;
		function<void(ll, ll, ll, ll)>dfs = [&](ll x, ll y, ll z, ll k)
			{
				ll ans = 1e18;
				if ((x && y) || k == len)
				{
					if (k != len) fg[k] = z, dfs(x, y, z, k + 1);
					else
					{
						ll jk = 0;
						for (ll i = 0; i < k; i++)
						{
							jk = jk * 10 + fg[i];
						}
						ll maxx = 0, minn = 1e18;
						ll pd = 0;
						for (ll i = 0; i < k; i++)
						{
							if (fg[i] > 0)pd = 1;
							if (pd == 0)continue;
							minn = min(minn, fg[i]);
							maxx = max(maxx, fg[i]);
						}
						//if (jk == 3343)
							//cout << z << endl;
						ans = maxx - minn;
						if (e > ans)e = ans, an = jk;
					}
					return;
				}
				else if (x)
				{
					ll u = r[k] - '0';
					if (u == z)fg[k] = u, dfs(1, 0, z, k + 1);
					else if (u > z) fg[k] = z, dfs(1, 1, z, k + 1);
					else
					{
						fg[k] = u, dfs(1, 0, z, k + 1);
						if (u != 0)
							fg[k] = u - 1, dfs(1, 1, z, k + 1);
					}
				}
				else if (y)
				{
					ll u = l[k] - '0';
					if (u == z) fg[k] = u, dfs(0, 1, z, k + 1);
					else if (u > z)
					{
						fg[k] = u, dfs(0, 1, u, k + 1);
						if (u != 9)
							fg[k] = u + 1, dfs(1, 1, z, k + 1);
					}
					else  fg[k] = z, dfs(1, 1, z, k + 1);
				}
				else
				{
					ll k1 = l[k] - '0';
					ll k2 = r[k] - '0';
					if (k1 == k2)
						fg[k] = k1, dfs(0, 0, z, k + 1);
					else if (z == k1)
					{
						fg[k] = z, dfs(0, 1, z, k + 1);
						if (z + 1 != k2)fg[k] = z + 1, dfs(1, 1, z, k + 1);
						else fg[k] = z + 1, dfs(1, 0, z, k + 1);
					}
					else if (z == k2)
					{
						fg[k] = z, dfs(1, 0, z, k + 1);
						if (z - 1 != k1)fg[k] = z - 1, dfs(1, 1, z, k + 1);
						else fg[k] = z - 1, dfs(0,1, z, k + 1);
					}
					else if (z < k1)
					{
						fg[k] = k1, dfs(0, 1, z, k + 1);
						if (k1 + 1 != k2)fg[k] = k1 + 1, dfs(1, 1, z, k + 1);
						else fg[k] = k1 + 1, dfs(1, 0, z, k + 1);
					}
					else if (z > k2)
					{
						fg[k] = k2, dfs(1, 0, z, k + 1);
						if (k2 - 1 != k1)fg[k] = k2 - 1, dfs(1, 1, z, k + 1);
						else fg[k] = k2 - 1, dfs(0, 1, z, k + 1);
					}
					else fg[k] = z, dfs(1, 1, z, k + 1);
				}
				return;
			};
		for (ll i = 0; i <= 9; i++)
		{
			dfs(0, 0, i, 0);
		}
		cout << an << endl;
	}
}