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 binarySearch
	public 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)/2
			rest=x-y*mid;
			if(rest<0){
				high=mid-1;
			}
			if(rest>=y){
				low=mid+1;
			}
		}while(!(rest>=0&&rest<y));
		return mid;
	}
	
}