打印 1...nX8+n=n...1

public class PrintTest{public static void main(String[] args){//题一打印如下//        1X8+1=9//       12X8+2=98//      123X8+3=987//     1234X8+4=9876//    12345X8+5=98765//   123456X8+6=987654//  1234567X8+7=9876543// 12345678X8+8=98765432//123456789X8+9=987654321//解题思路////1.基数//这里的基数就是等号前边的数//看起来跟行数没什么区别////2.起始空格//可以看出是8-0个，并可以根据9-行数//来计算，行数为1-9,也就是正好8-0个////3.乘数//乘数可以通过基数来求//那么这个乘数可以看成是这样的// 1*(10^base-1) + 2*(10^base-2) .... base*(10^base-base)////4.表达式的和//第一种用求出来的乘数*8+基数//第二种//根据乘数和最后的表达式和的每一位相加//都可以得到10，那么有如下表达式//10*(10^base-1) + 10*(10^base-2) .... 10*(10^base-base)//最后表达式的和可以通过这个表达式减去乘数表达式来求//最后的和////也可以通过两个表达式的最终相减来求算数表达式的和question1();}public static void question1(){//第一种解题方式采用字符方式打印for (int line = 1; line <= 9; line++)        {        printQuestion1_1(line);        }System.out.println("=======================>");//第二种解题方式采用字符串方式打印for (int line = 1; line <= 9; line++)        {        printQuestion1_2(line);        }System.out.println("=======================>");//第三种解题方式采用算法方式打印//1.通过基数计算出乘数//2.计算出表达式的和for (int line = 1; line <= 9; line++)        {        printQuestion1_3(line);        }System.out.println("=======================>");//第四种解题方式采用算法方式打印//从各位开始逐位计算乘数和表达式和的每一位//并将最后的结果相加for (int line = 1; line <= 9; line++)        {        printQuestion1_4(line);        }System.out.println("=======================>");//第五种解题方式采用算法方式打印//计算出乘数乘数和表达式每一位的和，也就是10//并最后用这个和减去乘数得到表达式的和for (int line = 1; line <= 9; line++)        {        printQuestion1_5(line);        }System.out.println("=======================>");//第六种解题方式跟第三种类似//计算乘数的方法不一样for (int line = 1; line <= 9; line++)        {        printQuestion1_6(line);        }System.out.println("=======================>");}public static void printQuestion1_1(int base){//计算本行有多少空格并打印for (int space = 9 - base; space > 0; space--)        {        System.out.print(" ");        }//计算本行的乘数，并打印for (int multiplier = 1; multiplier <= base; multiplier++)        {        System.out.print(multiplier);        }System.out.print("X8+"+base+"=");//计算表达式最后的和for (int sum = 9; sum > 9 - base; sum--)        {System.out.print(sum);        }System.out.println();}public static void printQuestion1_2(int base){//根据基数打印本行的空格String space = "         ";System.out.print(space.substring(base));//根据基数打印本行的乘数StringBuilder sb = new StringBuilder("123456789");System.out.print(sb.toString().substring(0,base));//打印倍数和技术的表达式    System.out.print("X8+" + base+"=");        //打印最后的和    System.out.print(sb.reverse().substring(0,base));    System.out.println();}public static void printQuestion1_3(int base){//计算本行有多少空格并打印for (int space = 9 - base; space > 0; space--)        {        System.out.print(" ");        }//计算本行的乘数并打印int multiplier = 0;for (int j = 0; j < base; j++)        {multiplier += j + 1;multiplier = j < base - 1 ? multiplier * 10 : multiplier;        }System.out.print(multiplier);//打印倍数和基数表达式int multiple = 8;System.out.print("X" + multiple + "+" + base + "=");//计算表达式最后的和并打印int sum = multiplier * 8 + base;System.out.print(sum);System.out.println();}public static void printQuestion1_4(int base){//计算本行有多少空格并打印for (int space = 9 - base; space > 0; space--)        {        System.out.print(" ");        }int multiplier = 0;int sum = 0;for (int i = base, factor = 1; i > 0; i--, factor *=10)        {multiplier+=(i*factor);sum+=((10-i)*factor);                }System.out.println(multiplier+"X8+"+base+"="+sum);}public static void printQuestion1_5(int base){//计算本行有多少空格并打印for (int space = 9 - base; space > 0; space--)        {        System.out.print(" ");        }int multiplier = 0;int sum = 10;for (int i = base, factor = 1; i > 0; i--, factor*=10)        {multiplier+=(i*factor);if (i != base)            {sum = sum*10+10;            }}System.out.println(multiplier+"X8+"+base+"="+(sum - multiplier));}public static void printQuestion1_6(int base){//计算本行有多少空格并打印for (int space = 9 - base; space > 0; space--)        {        System.out.print(" ");        }//乘数的求法，也可以这样//0*10+1 //0*10+1 + (0*10+1)*10+1//0*10+1 + (0*10+1)*10+1 + ((0*10+1)*10+1)*10+1//0*10+1 + (0*10+1)*10+1 + ((0*10+1)*10+1)*10+1 + (((0*10+1)*10+1)*10+1)*10+1//换换简单点的写法//1//1+11//1+11+111//1+11+111+1111//用命题的第九行减去第八行，第八行减去第七行//如此类推就能得到上边的数列//在根据要显示的行数把数列中的所有数相加就能//得到最后算数表达式的值int multiplier=0;for(int i = 0, factor = 0; i < base; i++){factor = factor * 10 + 1;multiplier += factor;}System.out.println(multiplier+"X8+"+base+"="+(multiplier*8+base));}}