标签:action sage egerp return [1] add stat integer 技术分享
一、请编写程序将整数转换为汉字读法字符串
1 import java.util.Scanner; 2 3 public class Test { 4 5 public static void main(String[] args) { 6 Scanner in=new Scanner(System.in); 7 System.out.println("输入一个数:"); 8 double a; 9 a=in.nextDouble(); 10 System.out.println(digitUppercase(a)); 11 } 12 public static String digitUppercase(double n) { 13 String fraction[] = { "角", "分"}; 14 String digit[] = { "零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖"}; 15 String unit[][] = { { "元", "万", "亿"}, { "", "拾", "佰", "仟"}}; 16 17 String head = n < 0 ? "负" : ""; 18 n = Math.abs(n); 19 20 String s = ""; 21 for (int i = 0; i < fraction.length; i++) { 22 s += (digit[(int) (Math.floor(n * 10 * Math.pow(10, i)) % 10)] + fraction[i]).replaceAll("(零.)+", ""); 23 } 24 if (s.length() < 1) { 25 s = "整"; 26 } 27 int integerPart = (int) Math.floor(n); 28 29 for (int i = 0; i < unit[0].length && integerPart > 0; i++) { 30 String p = ""; 31 for (int j = 0; j < unit[1].length && n > 0; j++) { 32 p = digit[integerPart % 10] + unit[1][j] + p; 33 integerPart = integerPart / 10; 34 } 35 s = p.replaceAll("(零.)*零$", "").replaceAll("^$", "零") + unit[0][i] + s; 36 } 37 return head + s.replaceAll("(零.)*零元", "元").replaceFirst("(零.)+", "").replaceAll("(零.)+", "零").replaceAll("^整$", "零元整"); 38 } 39 40 }
程序截图:
二、数组表示大数
1 import java.util.Scanner; 2 3 public class Dashu { 4 5 public static void main(String[] args) { 6 // TODO Auto-generated method stub 7 String s1,s2; 8 Scanner in=new Scanner(System.in); 9 System.out.println("输入第一个大数:"); 10 s1=in.next(); 11 System.out.println("输入第二个大数:"); 12 s2=in.next(); 13 System.out.println(add(s1,s2)); 14 System.out.println(sub(s1,s2)); 15 16 } 17 18 private static String add(String a, String b){ 19 System.out.print("加法:" + a+ "+" + b + "="); 20 char[] aa = new StringBuffer(a).reverse().toString().toCharArray(); 21 char[] bb = new StringBuffer(b).reverse().toString().toCharArray(); 22 int aLen = aa.length; 23 int bLen = bb.length; 24 25 int len = aLen > bLen ? aLen : bLen; 26 27 int[] result = new int[len + 1]; 28 for (int i = 0; i < len + 1; ++i) { 29 int aint = i < aLen ? aa[i] - ‘0‘ : 0; 30 int bint = i < bLen ? bb[i] - ‘0‘ : 0; 31 result[i] = aint + bint; 32 } 33 34 for(int i=0;i<result.length-1;++i){ 35 if(result[i]>=10){ 36 result[i+1] += result[i]/10; 37 result[i] %= 10; 38 } 39 } 40 41 boolean flag = true; 42 StringBuffer sb = new StringBuffer(len); 43 for(int i=len;i>=0;--i){ 44 if(result[i]==0&&flag){ 45 continue; 46 } 47 else{ 48 flag=false; 49 } 50 sb.append(result[i]); 51 } 52 return sb.toString(); 53 } 54 public static String sub(String f, String s){ 55 System.out.print("减法:" + f + "-" + s + "="); 56 // 将字符串翻转并转换成字符数组 57 char[] a = new StringBuffer(f).reverse().toString().toCharArray(); 58 char[] b = new StringBuffer(s).reverse().toString().toCharArray(); 59 int lenA = a.length; 60 int lenB = b.length; 61 int len = lenA > lenB ? lenA : lenB; 62 int[] result = new int[len]; 63 char sign = ‘+‘; 64 if (lenA < lenB) { 65 sign = ‘-‘; 66 } 67 else if (lenA == lenB){ 68 int i = lenA - 1; 69 while (i > 0 && a[i] == b[i]) { 70 i--; 71 } 72 if (a[i] < b[i]) { 73 sign = ‘-‘; 74 } 75 } 76 for (int i = 0; i < len; i++) { 77 int aint = i < lenA ? (a[i] - ‘0‘) : 0; 78 int bint = i < lenB ? (b[i] - ‘0‘) : 0; 79 if (sign == ‘+‘) { 80 result[i] = aint - bint; 81 } 82 else { 83 result[i] = bint - aint; 84 } 85 } 86 for (int i = 0; i < result.length - 1; i++) { 87 if (result[i] < 0) { 88 result[i + 1] -= 1; 89 result[i] += 10; 90 } 91 } 92 93 StringBuffer sb = new StringBuffer(); 94 if (sign == ‘-‘) { 95 sb.append(‘-‘); 96 } 97 boolean flag = true; 98 for (int i = len - 1; i >= 0; i--) { 99 if (result[i] == 0 && flag) { 100 continue; 101 } 102 else { 103 flag = false; 104 } 105 sb.append(result[i]); 106 } 107 if (sb.toString().equals("")) { 108 sb.append("0"); 109 } 110 System.out.println(sb.toString()); 111 return sb.toString(); 112 } 113 }
程序截图:
三、随机生成10个数,填充一个数组,然后用消息框显示数组内容,接着计算数组元素的和,将结果也显示在消息框中
1 import javax.swing.JOptionPane; 2 3 public class Shuzu { 4 5 public static void main(String[] args) { 6 // TODO Auto-generated method stub 7 int a[],sum=0; 8 a=new int[10]; 9 String output=new String(); 10 for(int i=0;i<a.length;i++){ 11 a[i]=(int)(Math.random()*100); 12 } 13 output+="a[10]={"; 14 for(int i=0;i<a.length;i++){ 15 output+=a[i]+" "; 16 } 17 output+="}"; 18 for(int i=0;i<a.length;i++){ 19 sum+=a[i]; 20 } 21 output+="\nsum:"+sum; 22 JOptionPane.showMessageDialog(null, output,"结果", 23 JOptionPane.PLAIN_MESSAGE); 24 } 25 }
程序截图:
设计思想:
随机产生10个相加数求和。
标签:action sage egerp return [1] add stat integer 技术分享
原文地址:http://www.cnblogs.com/XiaoPiHaiEr/p/6036598.html