零基础ACM学习的第一天

import java.util.Scanner;
public class Main {

public static void main(String[] args) {
	// TODO Auto-generated method stub
	Scanner sc=new Scanner(System.in);
	int t=sc.nextInt();
	System.out.print("L");
	for(int i=0;i<t;i++) {
		System.out.print("o");
	}
	System.out.print("ng");
	sc.close();
}

}
按题目要求，按部就班的写，（题目仅对o的个数有额外要求）。
import java.util.Scanner;
public class text1 {

public static void main(String[] args) {
	// TODO Auto-generated method stub
	Scanner sc=new Scanner(System.in);
	int t=sc.nextInt();
	sc.nextLine();
	for(int i=0;i<t;i++) {
		String a=sc.nextLine();
		a=a.toLowerCase();
		if(a.equals("yes")) {
			System.out.println("YES");
		}
		else {System.out.println("NO");}
	}
}

}
按题目要求，按部就班的写（先将输入都花成小写降低复杂性在比较）
import java.util.Scanner;
public class Main {

public static void main(String[] args) {
	// TODO Auto-generated method stub
	Scanner sc=new Scanner(System.in);
	int t=sc.nextInt();
	sc.nextLine();
	for(int i=0;i<t;i++) {
		String b=sc.nextLine();
		int a=b.charAt(b.length()-1)-'0';
		if(a%2==0) {
			System.out.println("even");
		}
		else {System.out.println("odd");}
	}sc.close();
}

}
只比较最后一个数是否为偶数
import java.util.Scanner;
public class Main {

public static void main(String[] args) {
	// TODO Auto-generated method stub
	Scanner sc=new Scanner(System.in);
	int t=sc.nextInt();
	for(int i=0;i<t;i++) {
		int n=sc.nextInt();
		int m=sc.nextInt();
		int A=0;
		int B=0;
		int C=0;
		int D=0;
		int E=0;
		int F=0;
		int G=0;
		int q=0;
		sc.nextLine();
		String a=sc.nextLine();
		for(int o=0;o<n;o++) {
			switch(a.charAt(o)) {
			case('A'):A++;break;
			case('B'):B++;break;
			case('C'):C++;break;
			case('D'):D++;break;
			case('E'):E++;break;
			case('F'):F++;break;
			case('G'):G++;break;
			}
		}
		if(A<m) {q+=m-A;}
		if(B<m) {q+=m-B;}
		if(C<m) {q+=m-C;}
		if(D<m) {q+=m-D;}
		if(E<m) {q+=m-E;}
		if(F<m) {q+=m-F;}
		if(G<m) {q+=m-G;}
		System.out.println(q);
	}
	sc.close();
}

}
统计各种难度题目出现的次数，和题目要求比较
import java.util.Scanner;
public class Main {

public static void main(String[] args) {
	// TODO Auto-generated method stub
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	int m=sc.nextInt();
	int k=sc.nextInt();
	int y=0;
	for(int o=0;o<m;o++) {
	int x=0;
	for(int i=0;i<n;i++) {
		int t=sc.nextInt();
		if(t==k) {x++;}
	}
	if(x>=(n+1)/2) {y++;}
}
	if(y>=(m+1)/2) {
		System.out.println("YES");
	}
	else {System.out.println("NO");
	}
	sc.close();
}

}
同过一个循环来求一天内是否同意，再来一个循环检测是否合理
import java.util.Scanner;

public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N, Q;
String S;

    N = sc.nextInt();
    S = sc.next();
    Q = sc.nextInt();

    char[] c = new char[26];

    for (int i = 0; i < 26; i++) {
        c[i] = (char) ('a' + i);
    }

    while (Q-- > 0) {
        char from = sc.next().charAt(0);
        char to = sc.next().charAt(0);

        for (int i = 0; i < 26; i++) {
            if (c[i] == from) {
                c[i] = to;
            }
        }
    }

    char[] sArray = S.toCharArray();
    for (int i = 0; i < sArray.length; i++) {
        if (sArray[i] >= 'a' && sArray[i] <= 'z') {
            sArray[i] = c[sArray[i] - 'a'];
        }
    }

    System.out.println(new String(sArray));
}

}
建立数组，先对操作历程进行处理，再处理输入的字符串
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.BufferedWriter;
import java.io.OutputStreamWriter;
import java.io.IOException;

public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
String[] firstLine = br.readLine().split(" ");
int a = Integer.parseInt(firstLine[0]);
int step = Integer.parseInt(firstLine[1]);
int[][] array = new int[a][a];
for (int i = 0; i < a; i++) {
String[] row = br.readLine().split(" ");
for (int j = 0; j < a; j++) {
array[i][j] = Integer.parseInt(row[j]);
}
}
int[] colMap = new int[a];
for (int i = 0; i < a; i++) {
colMap[i] = i;
}
for (int s = 0; s < step; s++) {
String[] opLine = br.readLine().split(" ");
int op = Integer.parseInt(opLine[0]);
int x = Integer.parseInt(opLine[1]);
int y = Integer.parseInt(opLine[2]);

        if (op == 1) { 
            int[] temp = array[x - 1];
            array[x - 1] = array[y - 1];
            array[y - 1] = temp;
        } else if (op == 0) { 
            int tempCol = colMap[x - 1];
            colMap[x - 1] = colMap[y - 1];
            colMap[y - 1] = tempCol;
        }
    }
    int[][] result = new int[a][a];
    for (int i = 0; i < a; i++) {
        for (int j = 0; j < a; j++) {
            result[i][j] = array[i][colMap[j]];
        }
    }
    for (int i = 0; i < a; i++) {
        StringBuilder sb = new StringBuilder();
        for (int j = 0; j < a; j++) {
            sb.append(result[i][j]).append(" ");
        }
        bw.write(sb.toString().trim());
        bw.newLine();
    }

    bw.flush();
    br.close();
    bw.close();
}

}
重点处理要得到的矩阵，同时用行映射和列映射，最后通过colMap的映射关系一次性构造最终结果矩阵。

学习总结：
时间复杂度的大小是代码运行速度的反应。如何改变算法，实现更小的时间复杂度简化代码。算法题目的与众不同的点让我刻骨铭心，总是修改自己的代码避免超时。C++还在进行初步的学习，auto的命名方式和以往的接触到的相比具有极大的便捷性。