Interview Notes---Logical Questions: Pour Water Problem

　　Calculate the optimal solution of the water pouring problem of the logic question of the interview:

    /**
     * Algorithms and principles
     * * / 
    public  int [] rsa ( int [] data) {
         / ** 
         * Logic question: There are two water cups, one is aL and the other is bL, water can be used casually, how to get cL
         * That is to use the principle of full in and out, either aL is full back to bL, or bL is full back to aL cycle to get
         * that is satisfied: ax-by=c/bx-ay=c
         * Regarding the number of times each cup is full, x/y is set to no more than 100 by default
         * */
        int a=data[0];
        int b=data[1];
        int c=data[2];
        
        int sum=0;
        for(int x=0;x<100;x++) {
            for(int y=0;y<100;y++) {
                if(a*x-b*y==c) {
                    sum=x+y;
                    //返回a,b,x,y,sum
                    int [] datas= {a,b,x,y,sum};
                    return datas;
                }
            }
        }
        return data;
    }
    
    /**
     * call the method,
     * Because there are two possibilities, that is, calling twice, the second time the ab order is exchanged and assigned
     * */ 
    public  void use( int [] data) {
         int a=data[0 ];
         int b=data[1 ];
         int c=data[2 ];
         // call 
        int [] data1= {a, b,c};
         int [] data2= {b,a,c};
         // Get the returned array 
        int [] s1 = rsa(data1);
         int [] s2= rsa(data2);
        
        // Judge the total number of steps and find the optimal solution 
        if (s1[4]<s2[4 ]) {
            String message=s1[0]+"升倒"+s1[1]+"升：x="+s1[2]+",y="+s1[3];
            System.out.println(message);    
        }else {
            String message=s2[0]+"升倒"+s2[1]+"升：x="+s2[2]+",y="+s2[3];
            System.out.println(message);    
        }
    }
}

　　Test example:

public static void main(String[] args) {
    dsUtils test = new dsUtils();
     // Pass the capacity of the two cups and the capacity to be taken 
    int [] data= {11,7,2 };
    test.use(data);
}

　　Test Results:

　　

　　If you don't need to find the optimal solution of the two methods, you don't need to be so troublesome:　

public  class dsUtil {
     /**
     * Algorithms and principles
     * */
    public void rsa(int a,int b,int c) {
        /** 
         * Logic question: There are two water cups, one is aL and the other is bL, water can be used casually, how to get cL
         * That is to use the principle of full in and out, either aL is full back to bL, or bL is full back to aL cycle to get
         * that is satisfied: ax-by=c/bx-ay=c
         * Regarding the number of times each cup is full, x/y is set to no more than 100 by default
         * */        
        for(int x=0;x<100;x++) {
            for(int y=0;y<100;y++) {
                if(a*x-b*y==c) {
                    String message=a+"升倒"+b+"升：x="+x+",y="+y;
                    System.out.println(message);
                    return;
                }
            }
        }    
    }
    
    /**
     * call the method,
     * Because there are two possibilities, that is, calling twice, the second time the ab order is exchanged and assigned
     * */
    public void use(int a,int b,int c) {
        rsa(a,b,c);
        rsa(b,a,c);
    }
}

View Code

 Test Results:

　　

　　Reference: https://www.cnblogs.com/j-star/p/7127190.html