HDOJ2074 叠筐

来源：互联网发布：广东11选5遗漏360数据编辑：程序博客网时间：2024/06/03 16:34
HDACM2074
改题需注意：输入的n，n/2%2的奇偶性，这是判断外环是前一个字符还是后一个字符的关键，博主以为是ASCII值小的在外面，被坑死了，但是没关系，坑坑更健康。
import java.util.Scanner;public class Main{    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        boolean boo = true;        while (sc.hasNext()) {            int n = sc.nextInt();            String str = sc.next();            char[] a = str.toCharArray();            str = sc.next();            char[] b = str.toCharArray();            if (boo) {                boo = false;            }else{                System.out.println();            }            if (n==1) {                System.out.println(a[0]);                continue;            }            char[][] pattern = new char[n/2+1][n];            if (n/2%2==0) {                for (int i = 0; i < pattern.length; i++) {                    for (int j = 0; j < pattern[i].length; j++) {                        if (i==0) {                            if (j == 0|| j==n-1) {                                pattern[i][j]=' ';                            }else {                                pattern[i][j]=a[0];                            }                        }else{                            if (j%2==0) {                                pattern[i][j]=a[0];                            }else{                                pattern[i][j]=b[0];                            }                            if(i%2==1){                                for (int j2 = i; j2 < n-i; j2++) {                                    pattern[i][j2]=b[0];                                }                                               }else {                                for (int j2 = i; j2 < n-i; j2++) {                                    pattern[i][j2]=a[0];                                }                            }                                           }                    }                }            }else {                for (int i = 0; i < pattern.length; i++) {                    for (int j = 0; j < pattern[i].length; j++) {                        if (i==0) {                            if (j == 0|| j==n-1) {                                pattern[i][j]=' ';                            }else {                                pattern[i][j]=b[0];                            }                        }else{                            if (j%2==0) {                                pattern[i][j]=b[0];                            }else{                                pattern[i][j]=a[0];                            }                            if(i%2==1){                                for (int j2 = i; j2 < n-i; j2++) {                                    pattern[i][j2]=a[0];                                }                                               }else {                                for (int j2 = i; j2 < n-i; j2++) {                                    pattern[i][j2]=b[0];                                }                            }                                           }                    }                }            }            for (int i = 0; i < pattern.length; i++) {                for (int j = 0; j < pattern[i].length; j++) {                    System.out.print(pattern[i][j]);                }                System.out.println();            }            for (int i = pattern.length-2; i >=0; i--) {                for (int j = 0; j < pattern[i].length; j++) {                    System.out.print(pattern[i][j]);                }                System.out.println();            }        }    }}
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