不用乘法或加法给一个数增加7倍

Question

如题

Answer 1

int x = 20;
x  = (x << 3) - x;

Answer 2

//不用乘法和加法给一个数增加7倍
public class DivideS
{
public static void main(String... args) {
//
int x = 8; 
int a = x << 2;
int b = x << 1;
int s = a^b^x;
System.out.println(s);
}
}

Answer 3

  i -= i<<3;
  System.out.println(-i);

Answer 4

(x << 3) - x

Answer 5

增加7倍就是变成8倍，左移3位即可：
x = x << 3;

Answer 6

x = x < < 3;

Answer 7

果然，
It's a trap!!!!!!

Answer 8

x = x << 3;最佳答案。我想没有更好的方法了吧。

Answer 9

//无符号不算最高位的加法,其它的也可以实现吧,减法等于加上它的补码,LZ要几倍都行public class Main {
    public static long sum(long a,long b){
        long c=0;
        long bf1=0x0001,bf=0;
        byte jw=0;
        for(int i=0;i<Long.SIZE;i++){
            long ai=a&(0x0001<<i);
            long bi=b&(0x0001<<i);
            if(ai>0){//1 //为了方便理解用if,不用?:运算
                if(bi>0){//1
                    if(jw>0){//1
                        jw=1;
                        bf|=0x0001<<i;//确定第i位为1,进1
                    }else{
                        jw=1;//确定第i位为0//进1
                    }
                }else{//0
                    if(jw>0){//1
                        jw=1;//确定第i位为0//进1
                    }else{
                        jw=0;
                        bf|=0x0001<<i;//确定第i位为1,进0
                    }
                }
            }else{//0
                if(bi>0){//1
                    if(jw>0){//1
                        jw=1;//确定第i位为0//进1
                    }else{
                        jw=0;
                        bf|=0x0001<<i;//确定第i位为1,进0
                    }
                }else{//0
                    if(jw>0){//1
                        jw=0;
                        bf|=0x0001<<i;//确定第i位为1,进0
                    }else{
                        jw=0;//确定第i位为0//进0
                    }
                }
            }
        }
        return bf;
    }
    public static void main(String[] args) throws FileNotFoundException {
        System.out.println(""+sum(20, 111));
    }
}

Answer 10

不用+运算实现加法
int Add(int a,int b)
{
int jw=a&b;
int jg=a^b;
while(jw)
{
int t_a=jg;
int t_b=jw<<1;//进位
jw=t_a&t_b;//求进位
jg=t_a^t_b;//求相加的结果，不考虑进位
}
return jg;
}

Answer 11

你不懂别乱说，这个其实是让人模拟cpu如何运算，搞清楚计算机如何运算

Answer 12

very excellent
learning~~~~~~~~~~~~~`