谨以此文巩固算法知识
本来只有六道题的，后来改成26道了，学业繁忙，暂时不写解析了，有时间会补充
题目一：九九乘法表
题目链接：OpenJudge – 001:九九乘法表
思路分析：
分析题目描述可知，共输出9行，每行的列由1至9递增，因此考虑双层嵌套循环
外层循环控制行数，内层循环控制列数
下面分析循环的控制条件
第一行有一列，第二行有二列，即行和列的关系为第n行有n列，即当外层循环到n的时候，内层循环要输出n次
所以只需外层循环从1循环到9，内层循环从1循环到当前外层循环的值即可，下面是代码实现
代码实现：
public class Main {public static void main(String[] args){for(int i=1;i<=9;i++){for (int j=1;j<=i;j++){System.out.printf(j+"*"+i+"="+j*i+"\t");}System.out.println();//这里是因为行与行间有换行符}}}题目二：斐波那契数列
题目链接：OpenJudge – 002:斐波那契数列
思路分析：
首先需要明确斐波那契数列的定义
如1,3,4,7,11
这种第1,2项为随机值，自第3项开始每项值为前两项之和的数列称为斐波那契数列
公式表示即为
题目给定的输入x,y,n分别为数列的第1,2项以及数列长度
要求的输出为以x第一项，y为第二项，总长度为n的数列
那么我们需要知道该数列的每一项的值
我们现在已知
即我们可以根据此公式求出该数列任一项的值
求法：
定义一个方法int fib(int x,int y,int n),该方法返回以x为第一项,y为第二项的数列的第n项的值
例如输入的n为5
那么有
而f(2)和f(1)的值是已知的，因此可以用递推的方法求出f(5),同理求出f(n),我们代码采用递归形式实现
int fib(int x,int y,int n){if(n==1)return x;if(n==2)return y;return fib(x,y,n-1)+fib(x,y,n-2);}现在我们可以求出该数列的任意一项，因此只需要从1循环至n，循环内调用fib方法即可输出该数列
代码实现：
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int x = sc.nextInt();int y = sc.nextInt();int n = sc.nextInt();for (int i = 1; i <= n; i++) {System.out.print(new Main().fib(x,y,i) + " ");}}public int fib(int x,int y,int n) {if (n == 1)return x;if (n == 2)return y;return fib(x,y,n - 1) + fib(x,y,n - 2);}}题目三：约瑟夫环问题
题目链接：OpenJudge – 003:约瑟夫问题
思路分析：
下面是我的错误想法，相信也是很多人的想法，不过误打误撞推导出了正确公式，仅供避雷参考
我们假设有7只猴子，数到3的猴子出圈，即n=7,m=3
如上图所示，画圈的编号表示出圈的猴子编号
第一次选出圈猴时，从1开始数，数到3，第二次选出圈猴时，应从4开始数，因此第二次选出圈猴时开头的编号为4，以此类推，最后剩下4为大王
根据上图，我们可以知道，第一次选出圈猴时，开头的编号必然为1，而大王的编号必然是上图中最后一行的开头的编号，因此我们只要找到行与行间编号的映射关系即可知道大王的编号
观察上图画圈的编号，3->6->2->7->5,对应关系为
[f(1)+m]%n=f(2)
[f(2)+m]%n=f(3)
即[f(n)+m]%n=f(n+1) 其中f(n)表示第n行对应的编号
可化为[f(n-1)+m]%n=f(n)
即f(n) = [f(n-1)+m]%n
这个公式是怎么得到的？
我们看第一行，编号3在第一个的第三个位置，下一行的第一个位置必然是(3+1)%7,那么下一行的第三个位置就是(3+1+2)%7,即（3+m）%n
此公式对任一列都适用，我们代入第一列试一试
f(1) = 1
f(2) = [f(1)+3]%7 = 4
f(3) =[f(2)+3]%7 = 0
可以看到，到f(3)这里出现了问题，由于取余符号的特性，7%7,等于0而不是7，这里出现了错误，因此需要修改一下公式，f(n) = [f(n-1)+m-1]%n+1，这样就避免了出现7%7这种情况
至此 我认为该问题已经解决，然而实际上每次出圈总人数都会发生变化，此公式并不适用于求每次出圈的编号即只对f(n)和f(1)适用
下面是明烛大佬给出的数学推导过程这或许是你能找到的最详细约瑟夫环数学推导！ – 知乎
对于不是单纯搞算法的我们来说，这些数学类知识只需要稍微理解记住即可，不必深究
公式为：f(n) = [f(n-1)+m-1]%n+1
已知f(1)=1,那么f(n)只需要递归即可
代码实现：
import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = 1, m = 1;while (n != 0 && m != 0) {n = sc.nextInt();m = sc.nextInt();if(n==0&&m==0)break;System.out.println(new Main().yuesefu(n, m));}}public int yuesefu(int n, int k) {if (n == 1) {return 1;} else {return (yuesefu(n - 1, k) + k - 1) % n + 1;}}}题目四：全排列
原题链接：OpenJudge – 004:全排列
思路分析：
首先需要知道字母序的定义，用大白话来说，就是先比较第一个字符， 小的排前面，如果第一个字符相同，则比较第二个字符，小的排前面，以此类推
题目要求字符串内所有字符的所有排列，我们首先想到的肯定是暴力循环，但是显然我们并不知道循环的层数，因此采用递归回溯的形式求解（递归实质也是暴力循环，但是可以通过结束条件控制循环层数）
import java.util.ArrayList;import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {static ArrayList array = new ArrayList();char result[] = new char[7];int result_index = 0;public static void main(String[] args) {Scanner sc = new Scanner(System.in);String str = sc.next();char a[] = str.toCharArray();new Main().backtrack(a, 0, new boolean[a.length]);for (int i = 0; i = a.length) {if (result_index == a.length) {String str = new String(result, 0, result_index);array.add(str);}return;}for (int i = 0; i < a.length; i++) {if (!used[i]) {result[result_index++] = a[i];used[i] = true;backtrack(a, k + 1, used);result_index--;used[i] = false;}}}}array是String类型的集合，用于存储所有的排列好的字符串（这里也可以用String数组，但是会造成不必要的空间浪费）
result用于接收每次递归的字符，当达到原字符串的长度时，转成String存储到array中
递归回溯实际上就是对树的遍历
我们假设有a,b,c三个字符，则其树的结构为
由上图可知在result中同一元素会重复，因此使用used数组的boolean值来控制是否添加该元素，元素进入result数组之后，其对应的used数组的值改为true,这样在这一条路径上该元素不会再次进入result数组，在回溯时将该元素的used值设置为false，这样可以在同一层级上的其他路径上再次进入result数组，至于具体代码实现过程，实在繁琐，不再叙述
代码实现：
import java.util.ArrayList;import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {static ArrayList array = new ArrayList();char result[] = new char[7];int result_index = 0;public static void main(String[] args) {Scanner sc = new Scanner(System.in);String str = sc.next();char a[] = str.toCharArray();new Main().backtrack(a, 0, new boolean[a.length]);for (int i = 0; i = a.length) {if (result_index == a.length) {String str = new String(result, 0, result_index);array.add(str);}return;}for (int i = 0; i < a.length; i++) {if (!used[i]) {result[result_index++] = a[i];used[i] = true;backtrack(a, k + 1, used);result_index--;used[i] = false;}}}}题目五：统计质数个数
原题链接：OpenJudge – 005:统计质数个数
思路分析：
首先需要知道质数（素数）的定义，大白话来讲，如果一个数除了可以整除自身外，没有其他可以整除的数，那么这个数就是质数，如2,3,5,7
那么我们要知道一个数是不是素数，只需要知道其除了自身，是否还可以整除其他数
如判断7是不是质数，我们可以从2开始遍历到6，如果之间存在7%i==0，即存在i可以被7整除，那么7就不是质数，如果不存在，则7是质数，显然(7%i!)=0 (i=2,3,4,5,6)，因此7是质数
判断8是不是质数，我们从2遍历到7，显然8%2==0，因此8不是质数
但是此方法时间复杂度较高，可以做适当优化，我们知道，一个数的因数一定是关于其二次根对称的，以16为例，sqrt(16)=4,16=2*8,16=4*4,16=8*2,我们可以观察到，2,8是关于4对称的，那么如果在4之前没有数可以被16整除，则在4-15也不会有数可以被16整除，因此我们只需要遍历到sqrt(n)即可
知道了怎么判断一个数为质数，那么此问题只需要再嵌套一层循环从n遍历到m
即可
代码实现：
import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int count = 0;while (n = 2) {int s = 0;for (int i = 2; i <= n / i; i++) {if (n % i == 0) {s++;break;}}if (s == 0)count++;}n++;}System.out.println(count);}}题目六：水仙花数
原题链接：OpenJudge – 006:水仙花数（daffodil）
思路分析：
首先明确什么是水仙花数，对于一个三位数abc，如果a^3+b^3+c^3=abc，那么该数就是水仙花数，如153
此题的考点在于对于一个三位数如何单独取出其个位数，十位数，百位数
如153，其个位数为153%10=3,
其十位数为（153/10）%10=5，
其百位数为（153/10/10）%10=1
知道这一点 这道题就基本完成了
代码实现：
import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();for (int i = 100; i <= n; i++) {if ((i % 10) * (i % 10) * (i % 10) + (i / 10 % 10) * (i / 10 % 10) * (i / 10 % 10) + (i / 100 % 10) * (i / 100 % 10) * (i / 100 % 10) == i)System.out.println(i);}}}题目七：八皇后问题
代码实现：
/** * @author Sane * @version 1.0 */public class Main {static int num = 1;public static void main(String[] args) {int a[] = new int[8];new Main().backTrack(a, 0);}void backTrack(int a[], int k) {if (k == 8) {System.out.println("No. " + num++);for (int i = 0; i < 8; i++) {for (int j = 0; j < 8; j++) {if (j == search(a, i))System.out.print("1 ");elseSystem.out.print("0 ");}System.out.println();}return;}for (int i = 0; i < 8; i++) {if (check(a, i, k)) {a[k] = i;backTrack(a, k + 1);a[k] = 0;}}}boolean check(int a[], int i, int k) {for (int j = 0; j < k; j++) {if (a[j] == i || Math.abs(a[j] - i) == Math.abs(j - k)) {return false;}}return true;}int search(int a[], int i) {for (int j = 0; j < 8; j++) {if (a[j] == i)return j;}return 0;}}题目八：数字三角形
代码实现：
import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int N = sc.nextInt();int[][] a = new int[N][N];for (int i = 0; i < N; i++) {for (int j = 0; j <= i; j++) {a[i][j] = sc.nextInt();}}int[][] dp = new int[N][N];dp[0][0] = a[0][0];for (int i = 1; i < N; i++) {dp[i][0] = dp[i-1][0] + a[i][0];for (int j = 1; j <= i; j++) {dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j-1]) + a[i][j];}}int max = 0;for (int i = 0; i < N; i++) {max = Math.max(max, dp[N-1][i]);}System.out.println(max);}}题目九：放苹果
代码实现：
import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int t = sc.nextInt();while (t-- > 0) {int M = sc.nextInt();int N = sc.nextInt();System.out.println(countWays(M, N));}}public static long countWays(int M, int N) {if (M <= 1 || N == 1) {return 1;}if (M < N) {return countWays(M, M);}return countWays(M, N - 1) + countWays(M - N, N);}}题目十：鸣人和佐助
代码实现：
import java.util.ArrayDeque;import java.util.Scanner;import java.util.Queue;/** * @author Sane * @version 1.0 */class Main {static int hang, lie, chark;static Queue que = new ArrayDeque();static char map[][];static xy mingren;static boolean visited[][][];public static void main(String[] args) {Scanner sc = new Scanner(System.in);hang = sc.nextInt();lie = sc.nextInt();chark = sc.nextInt();map = new char[hang + 2][lie + 2];visited = new boolean[hang + 1][lie + 1][chark + 1];for (int i = 1; i <= hang; i++) {String str = sc.next();for (int j = 1; j <= lie; j++) {map[i][j] = str.charAt(j - 1);}}for (int i = 1; i <= hang; i++) {for (int j = 1; j <= lie; j++) {if (map[i][j] == '@') {mingren = new xy(i, j, chark);break;}}}visited[mingren.x][mingren.y][chark] = true;bfs();}static void bfs() {int dx[] = {1, -1, 0, 0};int dy[] = {0, 0, -1, 1};que.add(mingren);while (!que.isEmpty()) {xy a = que.poll();for (int i = 0; i 0 && new_x 0 && new_y <= lie && !visited[new_x][new_y][a.charkla]) {switch (map[new_x][new_y]) {case '*':xy temp = new xy(new_x, new_y, a.charkla);temp.dist = a.dist + 1;visited[new_x][new_y][a.charkla] = true;que.add(temp);break;case '+':System.out.println(a.dist + 1);return;case '#':if (a.charkla != 0) {xy temp2 = new xy(new_x, new_y, a.charkla - 1);temp2.dist = a.dist + 1;visited[new_x][new_y][a.charkla] = true;que.add(temp2);}break;}}}}System.out.println("-1");}}class xy {int x, y, charkla;int dist;public xy(int x, int y, int charkla) {this.x = x;this.y = y;this.charkla = charkla;}}题目十一：电影节
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int n;static ArrayList arrays;public static void main(String[] args) {Scanner sc = new Scanner(System.in);arrays = new ArrayList();while ((n = sc.nextInt()) != 0) {while (n-- > 0) {arrays.add(new movie(sc.nextInt(), sc.nextInt()));}Collections.sort(arrays, new Comparator() {public int compare(movie m1, movie m2) {return m1.end - m2.end;}});int last_end = 0;int count = 0;for (movie m : arrays) {int start = m.start;int end = m.end;if (start>=last_end) {count++;last_end = end;}}System.out.println(count);arrays.clear();}}}class movie {int start;int end;public movie(int start, int end) {this.start = start;this.end = end;}}题目十二：数制转换
代码实现：
import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int a = sc.nextInt();String str = sc.next();int b = sc.nextInt();if (str.equals("0")) {System.out.println("0");return;}long ten_n = to_ten(str, a);String s_turn = to_b(ten_n, b);System.out.print(s_turn);}public static long to_ten(String str, int a) {char num[] = str.toCharArray();long ten_n = 0;int count = 0;for (int i = num.length - 1; i >= 0; i--) {if (num[i] >= 'A' && num[i] = 'a' && num[i] 0; i++) {long temp = ten_n % b;if (temp >= 0 && temp <= 9) {s_turn = (temp + "") + s_turn;} else {s_turn = (char) ('A' + (temp - 10)) + s_turn;}ten_n /= b;}return s_turn;}}题目十三：胡
代码实现
import java.util.*;class Main {static int A[] = new int[10];static ArrayList que = new ArrayList();public static void main(String[] args) {Scanner sc = new Scanner(System.in);String str;while (true) {str = sc.nextLine().replaceAll(" ", "");if (str.charAt(0) == '0')break;for (int i = 0; i < str.length(); i++) {A[(int) (str.charAt(i) - '0')]++;}if ((str.length() - 2) % 3 != 0) {System.out.println("XIANGGONG");Arrays.fill(A, 0);} else {for (int i = 0; i = 2)que.add(i);}if (que.isEmpty()) {System.out.println("BUHU");Arrays.fill(A, 0);} else {boolean s = true;while (!que.isEmpty()) {s = true;int temp = que.remove(0);int B[] = Arrays.copyOf(A, 10);B[temp] -= 2;for (int i = 0; i < 10; i++) {B[i] %= 3;}boolean turn = true;while (turn) {int count = 0;for (int j = 0; j 0) {turn = true;count++;if (count % 3 == 0) {B[j - 2]--;B[j - 1]--;B[j]--;count = 0;} else turn = false;} else count = 0;}}for (int i = 0; i < 10; i++) {if (B[i] != 0) {s = false;}}if (s) {System.out.println("HU");Arrays.fill(A, 0);que.clear();break;}}if (!s) {System.out.println("BUHU");}}}Arrays.fill(A, 0);que.clear();}}}题目十四：垂直直方图
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static char map[][];static int counts[] = new int[26];public static void main(String[] args) {Scanner sc = new Scanner(System.in);for (int i = 0; i < 4; i++) {String str = sc.nextLine();for (int j = 0; j = 'A' && str.charAt(j) <= 'Z')counts[str.charAt(j) - 'A']++;}}int max = 0;for (int i = 0; i < 26; i++) {max = Math.max(max, counts[i]);}for (int i = 0; i <= max; i++) {if (i == max)for (int j = 0; j < 26; j++) {System.out.print((char) ('A' + j));System.out.print(' ');}elsefor (int j = 0; j = max - counts[j])System.out.print("*");elseSystem.out.print(' ');System.out.print(' ');}System.out.println();}}}题目十五：算24
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static double a[] = new double[4];static boolean suc = false;public static void main(String[] args) {Scanner sc = new Scanner(System.in);boolean ctn = true;while (true) {suc = false;for (int i = 0; i 题目十六：Knight Moves
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int n;static boolean map[][];static boolean visited[][];static ArrayList que = new ArrayList();static int l;static int dist[][];public static void main(String[] args) {Scanner sc = new Scanner(System.in);n = sc.nextInt();while (n-- > 0) {l = sc.nextInt();map = new boolean[l][l];visited = new boolean[l][l];dist = new int[l][l];int startx = sc.nextInt();int starty = sc.nextInt();visited[startx][starty] = true;int endx = sc.nextInt();int endy = sc.nextInt();if (startx == endx && starty == endy) {System.out.println(0);continue;}map[endx][endy] = true;bfs(startx, starty);System.out.println(dist[endx][endy]);que.clear();}}static void bfs(int startx, int starty) {que.add(new xy(startx, starty));while (!que.isEmpty()) {xy temp = que.remove(0);int x = temp.x;int y = temp.y;if (x - 2 >= 0) {if (y - 1 >= 0 && !visited[x - 2][y - 1]) {visited[x - 2][y - 1] = true;dist[x - 2][y - 1] = dist[x][y] + 1;if (map[x - 2][y - 1]) {return;}que.add(new xy(x - 2, y - 1));}if (y + 1 = 0) {if (y - 2 >= 0 && !visited[x - 1][y - 2]) {visited[x - 1][y - 2] = true;dist[x - 1][y - 2] = dist[x][y] + 1;if (map[x - 1][y - 2]) {return;}que.add(new xy(x - 1, y - 2));}if (y + 2 < l && !visited[x - 1][y + 2]) {visited[x - 1][y + 2] = true;dist[x - 1][y + 2] = dist[x][y] + 1;if (map[x - 1][y + 2]) {return;}que.add(new xy(x - 1, y + 2));}}if (x + 1 = 0 && !visited[x + 1][y - 2]) {visited[x + 1][y - 2] = true;dist[x + 1][y - 2] = dist[x][y] + 1;if (map[x + 1][y - 2]) {return;}que.add(new xy(x + 1, y - 2));}if (y + 2 < l && !visited[x + 1][y + 2]) {visited[x + 1][y + 2] = true;dist[x + 1][y + 2] = dist[x][y] + 1;if (map[x + 1][y + 2]) {return;}que.add(new xy(x + 1, y + 2));}}if (x + 2 = 0 && !visited[x + 2][y - 1]) {visited[x + 2][y - 1] = true;dist[x + 2][y - 1] = dist[x][y] + 1;if (map[x + 2][y - 1]) {return;}que.add(new xy(x + 2, y - 1));}if (y + 1 < l && !visited[x + 2][y + 1]) {visited[x + 2][y + 1] = true;dist[x + 2][y + 1] = dist[x][y] + 1;if (map[x + 2][y + 1]) {return;}que.add(new xy(x + 2, y + 1));}}}}}class xy {int x, y;public xy(int x, int y) {this.x = x;this.y = y;}}题目十七：末位零
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int N;public static void main(String[] args) {Scanner sc = new Scanner(System.in);N = sc.nextInt();while (N-- > 0) {int count = 0;int m = sc.nextInt();while (m > 0) {count += m / 5;m /= 5;}System.out.println(count);}}}题目十八：二叉树
代码实现
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int m, n;static int sum = 0;public static void main(String[] args) {Scanner sc = new Scanner(System.in);while (true) {m = sc.nextInt();n = sc.nextInt();if (m == 0 && n == 0)break;sum = 0;num(m, n, 0);System.out.println(sum);}}static void num(int m, int n, int hang) {if (2 * m <= n) {hang++;num(2 * m, n, hang);} else {if (hang == 0) {sum = 1;} else {int a = 1;sum = 1;for (int i = 1; i = n)sum = sum + (n - m) + 1;elsesum = sum + (m + 2 * a - 1 - m) + 1;}}}}题目十九：大整数加法
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);String str1 = sc.next();String str2 = sc.next();ArrayList al1 = new ArrayList();ArrayList al2 = new ArrayList();boolean i1 = true;boolean i2 = true;boolean zero1 = true;boolean zero2 = true;for (int i = 0; i < str1.length(); i++) {if (str1.charAt(i) != '0')zero1 = false;}for (int i = 0; i < str2.length(); i++) {if (str2.charAt(i) != '0')zero2 = false;}if (zero1 && zero2) {System.out.println(0);return;}for (int i = 0; i < str1.length(); i++) {if (i1) {if ((str1.charAt(i) - '0') == 0)continue;elsei1 = false;}al1.add(str1.charAt(i) - '0');}for (int i = 0; i < str2.length(); i++) {if (i2) {if ((str2.charAt(i) - '0') == 0)continue;elsei2 = false;}al2.add(str2.charAt(i) - '0');}Collections.reverse(al1);Collections.reverse(al2);ArrayList al3 = new ArrayList();boolean jinwei = false;for (int i = 0; i = 10) {jinwei = true;}}if (al1.size() > al2.size()) {for (int i = al2.size(); i = 10) {jinwei = true;}}} else {for (int i = al1.size(); i = 10) {jinwei = true;}}}if (jinwei)al3.add(1);Collections.reverse(al3);for (int i = 0; i < al3.size(); i++) {System.out.print(al3.get(i));}}}题目二十：显示器
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static boolean shang[] = new boolean[10];static boolean zhong[] = new boolean[10];static boolean xia[] = new boolean[10];static boolean start_shang[] = new boolean[10];static boolean start_xia[] = new boolean[10];static boolean end_shang[] = new boolean[10];static boolean end_xia[] = new boolean[10];public static void main(String[] args) {Scanner sc = new Scanner(System.in);while (true) {int size = sc.nextInt();int show_num = sc.nextInt();if (show_num == 0 && size == 0)break;shang[1] = true;shang[4] = true;zhong[0] = true;zhong[1] = true;zhong[7] = true;xia[1] = true;xia[4] = true;xia[7] = true;start_shang[0] = true;start_shang[4] = true;start_shang[5] = true;start_shang[6] = true;start_shang[8] = true;start_shang[9] = true;start_xia[0] = true;start_xia[2] = true;start_xia[6] = true;start_xia[8] = true;end_shang[0] = true;end_shang[1] = true;end_shang[2] = true;end_shang[3] = true;end_shang[4] = true;end_shang[7] = true;end_shang[8] = true;end_shang[9] = true;end_xia[0] = true;end_xia[1] = true;end_xia[3] = true;end_xia[4] = true;end_xia[5] = true;end_xia[6] = true;end_xia[7] = true;end_xia[8] = true;end_xia[9] = true;String num_str = String.valueOf(show_num);int len = num_str.length();for (int i = 0; i < 2 * size + 3; i++) {for (int k = 0; k < len; k++) {int temp = num_str.charAt(k) - '0';if (i == 0 || i == size + 1 || i == 2 * size + 2)System.out.print(" ");else {if (i size + 1) {if (start_xia[temp])System.out.print("|");else System.out.print(" ");}}for (int j = 0; j < size; j++) {if (i == 0) {if (shang[temp]) {System.out.print(" ");} else System.out.print("-");} else if (i == size + 1) {if (zhong[temp]) {System.out.print(" ");} else System.out.print("-");} else if (i == 2 * size + 2) {if (xia[temp]) {System.out.print(" ");} else System.out.print("-");} else {System.out.print(" ");}}if (i == 0 || i == size + 1 || i == 2 * size + 2)System.out.print(" ");else {if (i size + 1) {if (end_xia[temp])System.out.print("|");else System.out.print(" ");}}System.out.print(" ");}System.out.println();}System.out.println();}}}题目二十一：棋盘问题
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int n, k;static char map[][];static int count = 0;static boolean now_map[][];public static void main(String[] args) {Scanner sc = new Scanner(System.in);while (true) {n = sc.nextInt();k = sc.nextInt();if (n == -1 && k == -1) {break;}map = new char[n + 1][n + 1];now_map = new boolean[n + 1][n + 1];if (k <= 0 || n <= 0) {System.out.println(0);continue;}for (int i = 1; i <= n; i++) {String str = sc.next();for (int j = 1; j = k) {count++;return;}if (ceng > n)return;for (int i = 1; i <= n; i++) {if (map[ceng][i] == '#' && check(ceng, i)) {now_map[ceng][i] = true;dfs(ceng + 1, pulled + 1);now_map[ceng][i] = false;}}dfs(ceng + 1, pulled);}static boolean check(int i, int j) {for (int l = 1; l < i; l++) {if (now_map[l][j])return false;}return true;}}题目二十二：金币
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int day;static int sum = 0;public static void main(String[] args) {Scanner sc = new Scanner(System.in);while (true) {day = sc.nextInt();if (day == 0)break;answer(day);System.out.println(day + " " + sum);sum = 0;}}static void answer(int day) {int jiange = 1;int last_jiange = jiange;int gold = 1;for (int i = 1; i <= day; i++) {sum += gold;jiange--;if (jiange == 0) {jiange = last_jiange + 1;last_jiange = jiange;gold++;}}}}题目二十三：英语数字转换器
代码实现：
import java.util.HashMap;import java.util.Map;import java.util.Scanner;/** * @author Sane * @version 1.0 */public class Main {public static void main(String[] args) {Map dict = new HashMap();dict.put("negative", -1);dict.put("zero", 0);dict.put("one", 1);dict.put("two", 2);dict.put("three", 3);dict.put("four", 4);dict.put("five", 5);dict.put("six", 6);dict.put("seven", 7);dict.put("eight", 8);dict.put("nine", 9);dict.put("ten", 10);dict.put("eleven", 11);dict.put("twelve", 12);dict.put("thirteen", 13);dict.put("fourteen", 14);dict.put("fifteen", 15);dict.put("sixteen", 16);dict.put("seventeen", 17);dict.put("eighteen", 18);dict.put("nineteen", 19);dict.put("twenty", 20);dict.put("thirty", 30);dict.put("forty", 40);dict.put("fifty", 50);dict.put("sixty", 60);dict.put("seventy", 70);dict.put("eighty", 80);dict.put("ninety", 90);dict.put("hundred", 100);dict.put("thousand", 1000);dict.put("million", 1000000);Scanner scanner = new Scanner(System.in);while (scanner.hasNextLine()) {String line = scanner.nextLine();if (line.length() == 0) {break;}long ans = 0;long t = 0;String[] words = line.split(" ");for (String word : words) {if (dict.containsKey(word)) {int value = dict.get(word);if (value == -1) {System.out.print("-");} else if (value == 100) {t *= value;} else if (value == 1000 || value == 1000000) {t *= value;ans += t;t = 0;} else {t += value;}}}ans += t;System.out.println(ans);}}}题目二十四：重建二叉树
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int index = 0;static String str1, str2;static ArrayList nodes = new ArrayList();public static void main(String[] args) {Scanner sc = new Scanner(System.in);while (true) {if (!sc.hasNext())break;str1 = sc.next();if (!sc.hasNext())break;str2 = sc.next();digui(str2, 0, null);bianli(nodes.get(0));System.out.println();nodes.clear();index = 0;}}static void digui(String str2, int dir, Node last_node) {if (str2.isEmpty())return;if (index == str1.length())return;char ch = str1.charAt(index);Node node = new Node(ch);nodes.add(node);index++;if (last_node != null) {if (dir == 0) {last_node.left = node;} else last_node.right = node;}String new_array[] = str2.split(String.valueOf(ch));if (new_array.length > 0) {digui(new_array[0], 0, node);}if (new_array.length > 1)digui(new_array[1], 1, node);}static void bianli(Node node) {if (node.left != null)bianli(node.left);if (node.right != null)bianli(node.right);System.out.print(node.data);}}class Node {char data;Node left;Node right;public Node(char data) {this.data = data;}}题目二十五：汉诺塔问题(Tower of Hanoi)
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int n;static int a = 1, b = 2, c = 3;static char ch[];public static void main(String[] args) {Scanner sc = new Scanner(System.in);n = sc.nextInt();ch = new char[3];for (int i = 0; i " + ch2);return;}tower(t1, t3, t2, num - 1, bianhao - 1);tower(t1, t2, t3, 1, bianhao);tower(t2, t1, t3, num - 1, bianhao - 1);}}题目二十六：数字方格
代码实现：
import java.util.*;/** * @author Sane * @version 1.0 */class Main {static int a1, a2, a3;static int n;static int max;public static void main(String[] args) {Scanner sc = new Scanner(System.in);n = sc.nextInt();for (int i = 0; i <= n; i++) {for (int j = 0; j <= n; j++) {for (int k = 0; k <= n; k++) {if ((i + j) % 2 == 0 && (j + k) % 3 == 0 && (i + j + k) % 5 == 0)max = Math.max(max, i + j + k);}}}System.out.println(max);}}