Codeforces Round ＃753 (Div. 3)

A. Linear Keyboard

Big water , Violence will do
Reference code ：

#include <iostream>
#include <unordered_map>
#include <cmath>
using namespace std;

const int N = 50;
unordered_map<char, int> q;

int main()
{
    
	string s1, s2;
	int T;
	cin >> T;
	
	while(T--)
	{
    
		cin >> s1 >> s2;
		for(int i = 0; i < s1.size(); i++)
			q[s1[i]] = i + 1;
		int ans = 0;
		for(int i = 1; i < s2.size(); i++)
		{
    
			if(s2[i] == s2[i - 1]) continue;
			ans += abs(q[s2[i]] - q[s2[i - 1]]);
		}
		
		cout << ans << endl;
	}
} 

B. Odd Grasshopper

Beat the watch to find the rules , It is found that the law of even numbers is ：- + + -
The law of odd numbers is ：+ - - +
Discuss in classification , Just take the remainder and find the answer
Reference code ：

#include <iostream>
#include <cstdio>
using namespace std;

typedef long long LL;

int main()
{
    
	int T;
	cin >> T;
	
	while(T--)
	{
    
		LL x, t;
		scanf("%lld%lld", &x, &t);
		
		if(x % 2 == 0)
		{
    
			if(t % 4 == 1) x -= t;
			else if(t % 4 == 2) x += 1;
			else if(t % 4 == 3) x += t + 1;
		}
		
		else
		{
    
			if(t % 4 == 1) x += t;
			else if(t % 4 == 2) x -= 1;
			else if(t % 4 == 3) x -= (t + 1);
		}
		
		printf("%lld\n", x);
	}
	return 0;
}

C. Minimum Extraction

A sorting problem , Prioritize , In the direct safety topic meaning simulation can
Reference code ：

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

typedef long long LL;
const int N = 2e5 + 10;

int a[N];
LL s[N];

int main()
{
    
	int T;
	cin >> T;
	while(T--)
	{
    
		int n;
		cin >> n;
		for(int i = 1; i <= n; i++) cin >> a[i];
		
		sort(a + 1, a + n + 1);
		
		LL ans = -2e9, sum = 0;
		for(int i = 1; i <= n; i++)
		{
    
			ans = max(ans, a[i] - sum);
			sum += (a[i] - sum);
		}
		
		cout << ans << endl;
	}
	return 0;
}

D. Blue-Red Permutation

I feel this question is a little difficult to think of , But when I think of it , It's very simple , Ah , Thinking is still not good ,
The former part , Use to subtract 1 Go and get , And corresponding in order of size
The latter part , Use to add 1 Go and get , And corresponding in order of size
Then enumerate and judge whether the corresponding number is feasible .
Reference code ：

#include <iostream>
#include <algorithm>
#include <vector> 
using namespace std;

int main()
{
    
	int T;
	cin >> T;
	
	while(T--)
	{
    
		int n;
		cin >> n;
		vector<int> a(n), b, r;
		for(int i = 0; i < n; i++) cin >> a[i];
		
		char ch;
		for(int i = 0; i < n; i++)
		{
    
			cin >> ch;
			if(ch == 'B') b.push_back(a[i]);
			else r.push_back(a[i]);
		}
		
		sort(b.begin(), b.end());
		sort(r.begin(), r.end());
		
		bool flag = true;  int t = 1;
		for(auto i : b)
		{
    
			if(i < t) flag = false;
			t++;
		}
		
		for(auto i : r)
		{
    
			if(i > t) flag = false;
			t++;
		}
		
		if(flag) puts("YES");
		else puts("NO");
	}
	
	return 0;
}