1.18

A

签到题，没啥可说的

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

const int N = 110;

int n,m;
int x[N],y[N];

void solve() {
	memset(x, 0, sizeof(x));
	memset(y, 0, sizeof(y));
	
	for(int i = 1; i <= n;i++) {
		cin>>x[i]>>y[i];
	}
	
	int sum1 = 0, sum2 = 0;
	for(int i = 1; i <= n; i++) {
		sum1 += x[i];
		x[i] = sum1;
		sum2 += y[i];
		y[i] = sum2;
	}
	
	pair<int,int>zuoshang;
	zuoshang.first = sum1;
	zuoshang.second = sum2 + m;
	pair<int,int>youxia;
	youxia.first = sum1 + m;
	youxia.second = sum2;
	
	int l = youxia.first - x[1];
	int r = zuoshang.second - y[1];
	int ans = l * 2 + r * 2;
	cout<<ans<<endl; 
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	
	
	int t;
	cin>>t;
	
	while(t--) {
		cin>>n>>m;
		//n 是次数，m是length
		solve(); 
	}
	
	return 0;	
}

B

• 网上有人说是拓扑排序，我这里是直接从小到大检索，有几个比他小就说明后面要多空几个位置。

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

const int N = 1e3 + 10;

int n;
int g[N][N];
string s[N];
int ans[N];
bool str[N];

void solve() {
	memset(str, false, sizeof(str));
	memset(g, 0, sizeof(g));
	memset(ans, 0, sizeof(ans));
	
	cin>>n;
	for(int i = 1; i <= n; i++) {
		cin>>s[i];
	}
	
	
	for(int i = 1; i <= n; i++) {
		for(int j = 0; j < i - 1; j++) {
			if(s[i][j] == '1') {
				s[i][j] = '0';
			}
		}
	}
	
	memset(str, false, sizeof(str));
	
	for(int i = 1; i <= n; i++) {
		int ant = 0;
		
		for(int j = 0; j < n; j++) {
			if(s[i][j] == '1') {
				ant++;
			}
		}
		
		int num = 0;
		
		for(int k = n  - ant + 1; k <= n; k++) {
			if(!str[k]) {
				num++;
			}
		}
		
		for(int k = n  - ant; k >= 1; k--) {
			if(!str[k] && num == ant) {
				str[k] = true;
				ans[k] = i;
				break;
			}

			if(!str[k]) {
				num++;
			}
		}
	}
//	1
//3
//000
//001
//010
//3 2 1
	for(int i = 1; i <= n ; i++) {
		cout<<ans[i]<<" ";
	}
	
	cout<<endl;
	
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	
	
	int t;
	cin>>t;
	
	while(t--) {
		solve();
	}
	
	return 0;	
}

C

• 简化问题后，就是只要我们保持最长为3，个数就是每增加一个字符，个数增加两个就ok
• 我们发现就是1 2 X X 1 2 会发现1 X 1或者2 X 2每插入一个数字就会长度为3的两个个回文串，只要保证我们插入的XX相异即可。

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

const int N = 1e3 + 10;

int n;
//个数增多，但是不能增长 

void solve() {
	cin>>n;
	if(n == 6) {
		cout<<"1 1 2 3 1 2"<<endl;
		return;
	}
	for(int i = 1; i <= n - 2; i++) {
		cout<<i<<" "; 
	}
	cout<<"1 2"<<endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	
	
	int t;
	cin>>t;
	
	while(t--) {
		solve();
	}
	
	return 0;	
}

搜索

import java.util.Scanner;

public class Main {
    static final int N = 110;
    static int n, m;
    static String[] s = new String[N];
    static int[][] g =  new int[N][N];
    static boolean[][] str = new boolean[N][N];
    static int[] dx = {1, 0, -1, 0};
    static int[] dy = {0, 1, 0, -1};
    
    public static void dfs(int x, int y) {
    	for(int i = 0; i < 4; i++) {
    		int nowx = x + dx[i];
    		int nowy = y + dy[i];
    		if(!str[nowx][nowy] && g[nowx][nowy] != 0) {
    			str[nowx][nowy] = true;
    			dfs(nowx, nowy);
    		}
    	}
    }
    
    public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		n = scanner.nextInt();
		m = scanner.nextInt();
		
		for(int i = 1; i <= n; i++) {
			s[i] = scanner.next();
		}
		
		for(int i = 1; i <= n; i++) {
			for(int j = 0; j < m; j++) {
				g[i][j + 1] = s[i].charAt(j) - '0';
			}
		}
		
		
		int ans = 0;
		
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= m; j++) {
				if(g[i][j] != 0 && !str[i][j]) {
					str[i][j] = true;
					dfs(i, j);
					ans++;
				}
			}
		}
		
		System.out.println(ans);
		
		return;
	}
}

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

const int N = 1e2 + 10;
int n,m;
string s[N];
int g[N][N];
bool str[N][N];
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

void dfs(int x, int y) {
	
	for(int i = 0; i < 4; i++) {
		int nowx = x + dx[i];
		int nowy = y + dy[i];
		if(!str[nowx][nowy] && g[nowx][nowy] != 0) {
			str[nowx][nowy] = true;
			dfs(nowx, nowy);
		}
	}
}


int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	
	cin>>n>>m;
	
	
	 
	for(int i = 1; i <= n; i++) {
		cin>>s[i];
	}
	
	for(int i = 1; i <= n; i++) {
		for(int j = 0; j < m; j++) {
			g[i][j + 1] = s[i][j] - '0';
		}
	}
	
	int ans = 0;
	
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) {
			if(g[i][j] != 0 && !str[i][j]) {
				str[i][j] = true;
				dfs(i, j);
				ans++;
			}
		}
	}
	
	cout<<ans<<endl;
	
	
	return 0;	
}