Algorithms 4th - 1.1 Basic Programming Model - EXERCISES
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1.1.1
a. 7
b. 200.0000002
c. true
1.1.2
a. 1.618
b. 10.0
c. true
d. 33
1.1.3
1 public class MainApp { 2 public static void main(String[] args) { 3 4 int a = Integer.parseInt(args[0]); 5 int b = Integer.parseInt(args[1]); 6 int c = Integer.parseInt(args[2]); 7 8 if(a == b && b == c) { 9 System.out.println("equal"); 10 } else { 11 System.out.println("not equal"); 12 } 13 } 14 }
1.1.4
a. 去掉"then"
b. a > b 外围加括号
c. 正确
d. else 之前加冒号
1.1.5
public class MainApp { public static void main(String[] args) { Double a = Double.parseDouble(args[0]); Double b = Double.parseDouble(args[1]); if(a >= 0 && a <= 1 && b >=0 && b <= 1) { System.out.println("true"); } else { System.out.println("false"); } } }
1.1.6
斐波那契数列
0 -> 1 -> 1 -> 2 -> 3 -> 5 -> 8 -> 13 -> 21 -> 34 -> 55 -> 89 -> 144 -> 233 -> 377 -> 610
1.1.7
a. 3.00009
b. 499500
c. 10000
1.1.8
a. b
b. bc
c. e
1.1.9
int N = 13; String s = ""; for(int i = N; i > 0; i /= 2) { s = i % 2 + s; }
1.1.10
数组没有初始化
1.1.11
public class TestApp1 { public static void main(String[] args) { int N = 10; boolean ma[][] = new boolean[N][N]; for(int i = 0; i < N; i++) { for(int j = 0; j < N; j++) { double r = Math.random(); if(r < 0.5) ma[i][j] = true; else ma[i][j] = false; } } for(int i = 0; i < N; i++) { for(int j = 0; j < N; j++) { if(ma[i][j] == true) System.out.print("*"); else System.out.print(" "); } System.out.println(); } } }
1.1.12
0 1 2 3 4 4 3 2 1 0
1.1.13
public class TestApp1 { public static void main(String[] args) { int N = 10; int M = 15; boolean ma[][] = new boolean[N][M]; for(int i = 0; i < N; i++) { for(int j = 0; j < M; j++) { double r = Math.random(); if(r < 0.5) ma[i][j] = true; else ma[i][j] = false; } } System.out.println("origin matrix:"); for(int i = 0; i < N; i++) { for(int j = 0; j < M; j++) { if(ma[i][j] == true) System.out.print("*"); else System.out.print(" "); } System.out.println(); } System.out.println("transposition matrix:"); for(int i = 0; i < M; i++) { for(int j = 0; j < N; j++) { if(ma[j][i] == true) System.out.print("*"); else System.out.print(" "); } System.out.println(); } } }
1.1.14
public class TestApp1 { public static void main(String[] args) { System.out.println(lg(15)); System.out.println(lg(16)); System.out.println(lg(17)); } private static int lg(int N) { int result = 1; while(result <= N) { result *= 2; } return result / 2; } }
1.1.15
public class TestApp1 { public static void main(String[] args) { int A[] = {1, 2, 3, 3, 2, 5, 5, 4}; int N = 8; int M[] = histogram(A, N); for(int m : M) { System.out.println(m); } } private static int[] histogram(int[] A, int N) { int[] M = new int[N]; for(int i = 0; i < N; i++) { M[A[i]]++; } return M; } }
1.1.16
311461142246
1.1.17
由于一直会递归调用，所以不会停止，直到栈溢出。
1.1.18
修改前：
mystery(2, 25) = 50
mystery(3, 11) = 33
修改后：
mystery(2, 25) = 33554432
mystery(3, 11) = 177147
1.1.19
public class TestApp1 { private static long[] list = new long[1000]; public static void main(String[] args) { F(1000); for(int i = 0; i < list.length; i++) { if(i > 0 && list[i] == 0) { break; } StdOut.println(i + " " + list[i]); } } public static void F(int N) { list[0] = 0; list[1] = 1; for(int i = 2; i < N; i++) { list[i] = list[i - 1] + list[i - 2]; if(list[i] < list[i - 1]) { list[i] = 0; System.out.println("MAX : " + list[i - 1]); break; } } } }
最大值为 7540113804746346429
1.1.20
public class TestApp1 { public static void main(String[] args) { for(int i = 1; i <= 10; i++) { System.out.println("log_fact(" + i + "): " + log_fact(i)); } } private static double log_fact(int n) { if(n == 1) return 0; else return log_fact(n - 1) + Math.log(n); } }
1.1.21
import java.util.ArrayList; import java.util.List; public class TestApp1 { public static void main(String[] args) { List<Info> list = new ArrayList<Info>(); String line; while(StdIn.hasNextLine()) { line = StdIn.readLine(); if(line.isEmpty()) { break; } String[] content = line.split(" "); Info info = new Info(content[0], Integer.parseInt(content[1]), Integer.parseInt(content[2])); list.add(info); } for(Info info : list) { StdOut.println(info); } } } class Info { private String name; private int x; private int y; public Info(String name, int x, int y) { this.name = name; this.x = x; this.y = y; } public String toString(){ return "1: " + name + "2: " + x + "3: " + y + "4: " + x * 1.0 / y; } }
1.1.22
import java.util.ArrayList; import java.util.List; public class TestApp1 { public static void main(String[] args) { rank(3, new int[]{1, 3, 5, 6, 8, 11, 25, 78, 345, 7653}); } private static int rank(int key, int[] a) { return rank(key, a, 0, a.length - 1, 0); } private static int rank(int key, int[] a, int lo, int hi, int depth) { if(lo > hi) return -1; int mid = (lo + hi) / 2; int loop = depth; while(loop != 0) { System.out.print(" "); loop--; } System.out.println("lo:" + lo + " hi:" + hi); if(key < a[mid]) return rank(key, a, lo, mid - 1, ++depth); else if(key > a[mid]) return rank(key, a, mid + 1, hi, ++depth); else return mid; } }
1.1.23
import java.util.ArrayList; import java.util.Arrays; public class BinarySearch { private BinarySearch(){} public static int rank(int key, int[] a) { int lo = 0; int hi = a.length - 1; while(lo <= hi) { int mid = (lo + hi) / 2; if(key < a[mid]) hi = mid - 1; if(key > a[mid]) lo = mid + 1; else return mid; } return -1; } public static void main(String[] args) { In in = new In(args[0]); int[] whitelist = in.readAllInts(); Arrays.sort(whitelist); String flag = args[1]; if("+".equals(flag)) { while(!StdIn.isEmpty()) { int key = StdIn.readInt(); if(rank(key, whitelist) == -1) { StdOut.println(key); } } }else if("-".equals(flag)) { while (!StdIn.isEmpty()) { int key = StdIn.readInt(); if(rank(key, whitelist) != -1) { StdOut.println(key); } } } else { StdOut.println("Error Flag"); } } }
1.1.24
public class GCD { public static void main(String[] args) { StdOut.println(gcd(105, 24)); } public static int gcd(int a, int b) { StdOut.println("a: " + a + " b: " + b); if(b == 0) return a; else return gcd(b, a % b); } }
1.1.25
设两数为a、b(b<a），用gcd(a,b）表示a，b的最大公约数，r=a mod b 为a除以b以后的余数，k为a除以b的商，即a÷b=k.......r。辗转相除法即是要证明gcd(a,b)=gcd(b,r）。
第一步：令c=gcd(a,b），则设a=mc，b=nc
第二步：根据前提可知r =a-kb=mc-knc=(m-kn)c
第三步：根据第二步结果可知c也是r的因数
第四步：可以断定m-kn与n互质【否则，可设m-kn=xd，n=yd，（d>1），则m=kn+xd=kyd+xd=(ky+x)d，则a=mc=(ky+x)dc，b=nc=ycd，故a与b最大公约数成为cd，而非c，与前面结论矛盾】
从而可知gcd(b,r)=c，继而gcd(a,b)=gcd(b,r）。
证毕。