java-编程实现两个正整数的除法，当然不能用除法操作符

public class MyDiv {/** * 题目：编程实现两个正整数的除法，当然不能用除法操作符。 * 方法1：除数不断乘以2，直到最接近被除数 * 方法2：二分查找 * 扩展题目：如何求出a%b的值，要求不能使用除法和求模运算！ * 解答：在上面求出商(假设为c)之后，a%b=a-(b*c); */private static boolean INVALID_INPUT;public static void main(String[] args) {int x=24;for(int y=1;y<=x;y++){System.out.printf("%d/%d=%d%n", x,y,div01(x,y));System.out.printf("%d/%d=%d%n", x,y,div02(x,y));}}// return x/y. Let y*2*2*2...to be close to x.public static int div01(int x, int y) {if (!(x > 0 && y > 0 && x >= y)) {INVALID_INPUT = true;return -1;}int result = 0;while (x >= y) {int f = 1;/* * may overflow,change it to (y*f)<=(x/2)  while (y * f * 2 <= x) { f = f * 2;  } */while (y * f <= (x >> 1)) {f = f * 2;}result += f;x -= f * y;}return result;}// return x/y. Like binarySearchpublic static int div02(int x, int y) {if (!(x > 0 && y > 0 && x >= y)) {INVALID_INPUT = true;return -1;}int low=1;int high=x;int mid=0;int rest=x;do{mid=(low&high)+(low^high)/2;//low+(high-low)/2rest=x-y*mid;if(rest<0){high=mid-1;}if(rest>=y){low=mid+1;}}while(!(rest>=0&&rest<y));return mid;}}