不使用除法实现除法运算 和模运算

代码如下。包含测试代码

原理就是：枚举二进制余数中1的位置 +二分搜索。O(32*32) 

不过此算法可以加上两重的二分搜索。时间复杂度会更优。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <time.h>
using namespace std;
void module(int a ,int b ,int *c,int*d)
{
    if (a < b)
    {
        *c = 0;
        *d  = a;
        return ;
    }
    unsigned int k = 0;
    int t  = 32;
    for (  int j = 0; j< t;j++)
    {
        int d = (a - b*(k+(1<<j)));
        if ( d < 0)
        {
            if (j ==0)
            {
                break;
            }
            k = k+(1<<(j-1));
            t = j;
            j = -1;//j++
        }
        if (d == 0)
        {
            k = k+(1<<j);
            break;
        }     
    }
    *c = k;
    *d = a-b*k;
}
void main()
{
    
    int i = 10000;
    while(i--)
    {
        srand(time(NULL));
        //int a = rand(),b = rand();
        int a = 1,b = 1;
        int c,d;
        
        module(a,b,&c,&d);
        
        if (c!=a/b || d !=a%b )
        {
            printf("%d %d  %d %d %d %d\n",a,b,c,d,a/b,a%b);
        }
        //printf("finished % d\n",i );
    }
    printf("finished\n" );
}

 另附上同样解法的简洁的代码

int div1(const int x, const int y) {
    int left_num = x;
    int result = 0;
    while (left_num >= y) {
        int multi = 1;
        while (y * multi <= (left_num >> 1)) 
        {
            multi = multi << 1;
        }
        result =result +multi;
        left_num = left_num - y*multi;
    }
    return result;
}
//优化
int div2(const int x, const int y) {
    unsigned int left_num = x;
    int result = 0;
    unsigned int t = 0x80000000;
    int move = 0;
    while(t)
    {
        if(t<=left_num)
        {
            result = result+t;
            left_num = left_num - t;
        }
        t = t>>1;
    }
    return result;
}