结构之法 算法之道 面试题 9、编程实现两个正整数的除法 编程实现两个正整数的除法,当然不能用除法操作符。

微软面试题:编程实现两个正整数的除法,当然不能用除法操作符。 思路比较简单,效率似乎并不太理想,先放在这里,以后再改良. #include using namespace std; int Div1(int x, int y)//若x >>> y(远大于)时,效率差 { int ans = 0; while(x >= y) { x -= y; ++ans; } return ans; 阅读详情


微软面试题  

编程实现两个正整数的除法

编程实现两个正整数的除法,当然不能用除法操作符。
// return x/y.
int div(const int x, const int y) 
{
  ....

}


答案解析 

类似折半查找

常人会从1,2,3,4,5,6。。。。n。。。。这么一直试下去,发现=效率很低,为什么不学学折半查找呢,查找商值,

我会这么做

(1)先查找出一个商的范围,这个跳转很快的,我个人认为 O(log 正无穷 )

我设的范围 是 (1,2),(3,6),(7,14),(15,30)。。。。。。。。

直到找到 (m,n), m <= 商值 <= n )

(2) 折半查找就好了 O(log n)

总结  总的时间复杂度为O( log 正无穷 +log n)


实验数据


experiment result(文本,这次没有上传照片,太长了,不好截图,请谅解)

请输入正整数 x,y
23421543534
34
x = 1946707054, y = 34,d = 1
x = 1946707054, y = 34,d = 2
x = 1946707054, y = 34,d = 2
x = 1946707054, y = 34,d = 3
x = 1946707054, y = 34,d = 6
x = 1946707054, y = 34,d = 6
x = 1946707054, y = 34,d = 7
x = 1946707054, y = 34,d = 14
x = 1946707054, y = 34,d = 14
x = 1946707054, y = 34,d = 15
x = 1946707054, y = 34,d = 30
x = 1946707054, y = 34,d = 30
x = 1946707054, y = 34,d = 31
x = 1946707054, y = 34,d = 62
x = 1946707054, y = 34,d = 62
x = 1946707054, y = 34,d = 63
x = 1946707054, y = 34,d = 126
x = 1946707054, y = 34,d = 126
x = 1946707054, y = 34,d = 127
x = 1946707054, y = 34,d = 254
x = 1946707054, y = 34,d = 254
x = 1946707054, y = 34,d = 255
x = 1946707054, y = 34,d = 510
x = 1946707054, y = 34,d = 510
x = 1946707054, y = 34,d = 511
x = 1946707054, y = 34,d = 1022
x = 1946707054, y = 34,d = 1022
x = 1946707054, y = 34,d = 1023
x = 1946707054, y = 34,d = 2046
x = 1946707054, y = 34,d = 2046
x = 1946707054, y = 34,d = 2047
x = 1946707054, y = 34,d = 4094
x = 1946707054, y = 34,d = 4094
x = 1946707054, y = 34,d = 4095
x = 1946707054, y = 34,d = 8190
x = 1946707054, y = 34,d = 8190
x = 1946707054, y = 34,d = 8191
x = 1946707054, y = 34,d = 16382
x = 1946707054, y = 34,d = 16382
x = 1946707054, y = 34,d = 16383
x = 1946707054, y = 34,d = 32766
x = 1946707054, y = 34,d = 32766
x = 1946707054, y = 34,d = 32767
x = 1946707054, y = 34,d = 65534
x = 1946707054, y = 34,d = 65534
x = 1946707054, y = 34,d = 65535
x = 1946707054, y = 34,d = 131070
x = 1946707054, y = 34,d = 131070
x = 1946707054, y = 34,d = 131071
x = 1946707054, y = 34,d = 262142
x = 1946707054, y = 34,d = 262142
x = 1946707054, y = 34,d = 262143
x = 1946707054, y = 34,d = 524286
x = 1946707054, y = 34,d = 524286
x = 1946707054, y = 34,d = 524287
x = 1946707054, y = 34,d = 1048574
x = 1946707054, y = 34,d = 1048574
x = 1946707054, y = 34,d = 1048575
x = 1946707054, y = 34,d = 2097150
x = 1946707054, y = 34,d = 2097150
x = 1946707054, y = 34,d = 2097151
x = 1946707054, y = 34,d = 4194302
x = 1946707054, y = 34,d = 4194302
x = 1946707054, y = 34,d = 4194303
x = 1946707054, y = 34,d = 8388606
x = 1946707054, y = 34,d = 8388606
x = 1946707054, y = 34,d = 8388607
x = 1946707054, y = 34,d = 16777214
x = 1946707054, y = 34,d = 16777214
x = 1946707054, y = 34,d = 16777215
x = 1946707054, y = 34,d = 33554430
x = 1946707054, y = 34,d = 33554430
x = 1946707054, y = 34,d = 33554431
x = 1946707054, y = 34,d = 67108862
x = 1946707054, y = 34,d = 67108862
x = 1946707054, y = 34,d = 50331646
x = 1946707054, y = 34,d = 50331646
x = 1946707054, y = 34,d = 50331647
x = 1946707054, y = 34,d = 67108861
x = 1946707054, y = 34,d = 67108861
x = 1946707054, y = 34,d = 58720254
x = 1946707054, y = 34,d = 58720254
x = 1946707054, y = 34,d = 58720254
x = 1946707054, y = 34,d = 50331648
x = 1946707054, y = 34,d = 58720253
x = 1946707054, y = 34,d = 58720253
x = 1946707054, y = 34,d = 54525950
x = 1946707054, y = 34,d = 54525950
x = 1946707054, y = 34,d = 54525951
x = 1946707054, y = 34,d = 58720252
x = 1946707054, y = 34,d = 58720252
x = 1946707054, y = 34,d = 56623101
x = 1946707054, y = 34,d = 56623101
x = 1946707054, y = 34,d = 56623102
x = 1946707054, y = 34,d = 58720251
x = 1946707054, y = 34,d = 58720251
x = 1946707054, y = 34,d = 57671676
x = 1946707054, y = 34,d = 57671676
x = 1946707054, y = 34,d = 57671676
x = 1946707054, y = 34,d = 56623103
x = 1946707054, y = 34,d = 57671675
x = 1946707054, y = 34,d = 57671675
x = 1946707054, y = 34,d = 57147389
x = 1946707054, y = 34,d = 57147389
x = 1946707054, y = 34,d = 57147390
x = 1946707054, y = 34,d = 57671674
x = 1946707054, y = 34,d = 57671674
x = 1946707054, y = 34,d = 57409532
x = 1946707054, y = 34,d = 57409532
x = 1946707054, y = 34,d = 57409532
x = 1946707054, y = 34,d = 57147391
x = 1946707054, y = 34,d = 57409531
x = 1946707054, y = 34,d = 57409531
x = 1946707054, y = 34,d = 57278461
x = 1946707054, y = 34,d = 57278461
x = 1946707054, y = 34,d = 57278461
x = 1946707054, y = 34,d = 57147392
x = 1946707054, y = 34,d = 57278460
x = 1946707054, y = 34,d = 57278460
x = 1946707054, y = 34,d = 57212926
x = 1946707054, y = 34,d = 57212926
x = 1946707054, y = 34,d = 57212927
x = 1946707054, y = 34,d = 57278459
x = 1946707054, y = 34,d = 57278459
x = 1946707054, y = 34,d = 57245693
x = 1946707054, y = 34,d = 57245693
x = 1946707054, y = 34,d = 57245694
x = 1946707054, y = 34,d = 57278458
x = 1946707054, y = 34,d = 57278458
x = 1946707054, y = 34,d = 57262076
x = 1946707054, y = 34,d = 57262076
x = 1946707054, y = 34,d = 57262076
x = 1946707054, y = 34,d = 57245695
x = 1946707054, y = 34,d = 57262075
x = 1946707054, y = 34,d = 57262075
x = 1946707054, y = 34,d = 57253885
x = 1946707054, y = 34,d = 57253885
x = 1946707054, y = 34,d = 57253886
x = 1946707054, y = 34,d = 57262074
x = 1946707054, y = 34,d = 57262074
x = 1946707054, y = 34,d = 57257980
x = 1946707054, y = 34,d = 57257980
x = 1946707054, y = 34,d = 57257980
x = 1946707054, y = 34,d = 57253887
x = 1946707054, y = 34,d = 57257979
x = 1946707054, y = 34,d = 57257979
x = 1946707054, y = 34,d = 57255933
x = 1946707054, y = 34,d = 57255933
x = 1946707054, y = 34,d = 57255934
x = 1946707054, y = 34,d = 57257978
x = 1946707054, y = 34,d = 57257978
x = 1946707054, y = 34,d = 57256956
x = 1946707054, y = 34,d = 57256956
x = 1946707054, y = 34,d = 57256956
x = 1946707054, y = 34,d = 57255935
x = 1946707054, y = 34,d = 57256955
x = 1946707054, y = 34,d = 57256955
x = 1946707054, y = 34,d = 57256445
x = 1946707054, y = 34,d = 57256445
x = 1946707054, y = 34,d = 57256445
x = 1946707054, y = 34,d = 57255936
x = 1946707054, y = 34,d = 57256444
x = 1946707054, y = 34,d = 57256444
x = 1946707054, y = 34,d = 57256190
x = 1946707054, y = 34,d = 57256190
x = 1946707054, y = 34,d = 57256190
x = 1946707054, y = 34,d = 57255937
x = 1946707054, y = 34,d = 57256189
x = 1946707054, y = 34,d = 57256189
x = 1946707054, y = 34,d = 57256063
x = 1946707054, y = 34,d = 57256063
x = 1946707054, y = 34,d = 57256064
x = 1946707054, y = 34,d = 57256188
x = 1946707054, y = 34,d = 57256188
x = 1946707054, y = 34,d = 57256126
x = 1946707054, y = 34,d = 57256126
x = 1946707054, y = 34,d = 57256126
x = 1946707054, y = 34,d = 57256065
x = 1946707054, y = 34,d = 57256125
x = 1946707054, y = 34,d = 57256125
x = 1946707054, y = 34,d = 57256095
x = 1946707054, y = 34,d = 57256095
x = 1946707054, y = 34,d = 57256095
x = 1946707054, y = 34,d = 57256066
x = 1946707054, y = 34,d = 57256094
x = 1946707054, y = 34,d = 57256094
x = 1946707054, y = 34,d = 57256080
x = 1946707054, y = 34,d = 57256080
x = 1946707054, y = 34,d = 57256081
x = 1946707054, y = 34,d = 57256093
x = 1946707054, y = 34,d = 57256093
x = 1946707054, y = 34,d = 57256087
x = 1946707054, y = 34,d = 57256087
x = 1946707054, y = 34,d = 57256088
x = 1946707054, y = 34,d = 57256092
x = 1946707054, y = 34,d = 57256092
x = 1946707054, y = 34,d = 57256090
x = 1946707054, y = 34,d = 57256090
x = 1946707054, y = 34,d = 57256090
x = 1946707054, y = 34,d = 57256089
商=57256089
请按任意键继续. . .


代码实现

test_01.cpp

// test_01.cpp : Defines the entry point for the console application.
//

#include<stdio.h>
#include<stdlib.h> 

int main()
{
	int half(int m,int n,int x,int y);
	int x,y,result;


	printf("请输入正整数 x,y \n");
	scanf("%d",&x);
	scanf("%d",&y);
	result=half(1,2,x,y);
	printf("商=%d \n",result) ;
	system("pause");
	return 0;
}


// 类似折半查找商值:running time(log n)
/* functiong check()
input(m,n,x,y)
out result

if( m <=result<=n )
   折半查找就ok
else return half( n+1,2*n+2 ,x,y )
*/
int half(int m,int n,int x,int y)
{
	int check(int x,int y, int d);

	if(check(x,y,m)==0 )
	{
		return m ;
	}
	else if( check(x,y,n)==0 )
	{
		return n;
	}
    else if( check(x,y,n) == -1 )
	{
		if( check(x,y,(n+m)/2) == 0 )
			return (n+m)/2;
		else if( check(x,y,(n+m)/2) == 1 )
			return half( (n+m)/2+1,n-1,x,y );
		else if( check(x,y,(n+m)/2) == -1 )
			return half( m+1,(n+m)/2-1,x,y );
	}
    else return half( n+1,2*n+2,x,y );

}


// check the relation of the right result and d :running time(1)
/* functiong check()
input(x,y,d)
if( d==result )
   return 0;
else if( d>result )
   return -1;
else return 1;
*/

int check(int x,int y, int d)
{
	printf( "x = %d, y = %d,d = %d\n",x,y,d );
	if( x-d*y<y && x-d*y>=0 )
	{		
		return 0 ;
	}
	else if( x-d*y>=y )
	{
		return 1;
	}
	else if( x-d*y< 0 )
	{
		return -1;
	}
}


上面包含不严谨之处,前几天经过有心善良之人提醒,于2017-6-16 16:34 更新

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<limits.h>

//#define INT_MAX 0x7fffffff


unsigned int half(unsigned int m,unsigned int n,unsigned int x,unsigned int y);
unsigned int check(unsigned int x,unsigned int y, unsigned int d);
unsigned int max(unsigned int a, unsigned int b);
unsigned int min(unsigned int a, unsigned int b);


unsigned int main(unsigned int argc, char *argv[])
//unsigned int main()
{
    unsigned int x, y, result;
    char** left;
    printf("正整数 x = %s, y = %s \n", argv[1], argv[2]);
    x = atoi(argv[1]);
    y = atoi(argv[2]);
    result=half(1,y,x,y);
    printf("商=%u \n", result);
    return 0;
}


unsigned int half(unsigned int m,unsigned int n,unsigned int x,unsigned int y)
{
    unsigned int dis;
    printf( "m = %u, n = %u, x = %u, y = %u, check m = %u, check n = %u \n", m,n,x,y, check(x, y, m), check(x, y, n));
    if(check(x,y,m)==-1){
        return 0;
    }
    else if(check(x,y,m)==0 )
    {
        return m ;
    }
    else if( check(x,y,n)==0 )
    {
        return n;
    }
    else if( check(x,y,n) == -1 )
    {
        dis = sqrt(n-m);
        dis = dis * sqrt(dis);
        return max(half(m,m+dis,x,y), half(m+dis+1,n,x,y));
    }
    else return half( n+1, min(INT_MAX ,pow(n+1, 2)), x, y );
}

// max : max of a and b

unsigned int max(unsigned int a, unsigned int b)
{
    return a > b ? a : b ;
}

// min : min of a and b
unsigned int min(unsigned int a, unsigned int b)
{
    return a > b ? b : a ;
}

// check: d 和 正确商的距离
unsigned int check(unsigned int x,unsigned int y, unsigned int d)
{
    if( y*d <= x && y*(d+1) > x)
    {
        return 0 ;
    }
    else if( y*(d+1) <= x )
    {
        return 1;
    }
    else if(y*d > x)
    {
        return -1;
    }
}


Win10本地账户密码遗忘?巧用指纹/PIN快速重置(家庭版/专业版通用指南) 本文提供了一份详细的Win10本地账户密码遗忘自救指南,适用于家庭版和专业版用户。核心方法是利用已登录的指纹或PIN码权限,通过创建新的管理员账户或使用系统工具,安全重置遗忘的本地账户密码,无需第三方工具或重装系统,有效保护个人数据。 阅读详情

相关推荐

php对接建行h5网页支付

php对接建行h5网页支付前言H5页面支付H5页面支付回调通知退款及其他的业务 前言 最近接到公司乙方要求,需要再支付上面增加建行的支付渠道。之前一直用的都是微信和支付宝这种第三方支付平台,银行直接对接比较少。没办法就只能上网搜索,搜索一番返现网上对接建行的资料比较少,没办法只好摸着石头过河,中间也借鉴了一些大佬的文章内容。 H5页面支付 根据建行的h5网页支付对接文档,还是可以看出是比较简单的。...

lihua100862的博客 5411

编程实现两个正整数的除法(不能用除法操作符)

/* 编程实现两个正整数的除法(不能用除法操作符)。 //*/

1616

FUXA实战:5步搞定基于Web的SCADA系统搭建(含Node.js版本避坑指南)

本文提供了一份基于开源工具FUXA搭建Web版SCADA系统的实战指南。文章重点解析了从环境准备到生产部署的五个核心步骤,并特别强调了Node.js版本选择对西门子S7 PLC通信兼容性的关键影响,帮助开发者避开常见陷阱,快速构建工业监控界面。

weixin_29172963的博客 140

编程实现两个正整数的除法,当然不能用除法操作符

<br />/* 编程实现两个正整数的除法,当然不能用除法操作符。 */ #include <stdio.h> int div_int(const int x, const int y) { int left_num, result; int multi; left_num = x; result = 0; multi = 0; while(left_num>=y) { multi = 1;

albon arith 4126

室内无人机自主定位:融合Mid360激光雷达与光流传感器的无GPS导航实践

本文详细介绍了室内无人机自主定位技术,通过融合Mid360激光雷达与光流传感器实现无GPS导航。文章分析了技术挑战、硬件配置、算法优化及实战调试,展示了在复杂室内环境中如何实现厘米级定位精度,为仓库巡检、隧道勘测等场景提供可靠解决方案。

weixin_30500105的博客 532

o型圈沟槽设计_O型密封圈——详细选型过程及步骤

一、密封圈常用材料二、密封圈常用密封形式密封圈分为活塞密封,活塞杆密封,端面密封,旋转密封。三、O型密封圈的选型规则O型密封圈的尺寸图径向安装O型圈选型(静态和动态运动)活塞杆密封选择适当规格的O形圈时,应该保证其外径(d1+2*d2)至少等于或者大于沟槽外径d6;当选择的O形圈直径d1大于250 mm时, O形圈外径将不能超过沟槽外径的3%;当选择的O形圈直径d1小于250mm时, O形圈外径将...

weixin_39856630的博客 2万+

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 I

zjcheer_up的专栏 2857

中国土地市场网爬取

实习期间老板需要对土地数据进行爬取,想来想去就Google到了.应该算是种类最齐全的土地数据了。Github地址:https://github.com/AnTi-anti/china_land/tree/master。

猫敷雪 5462

编程实现两个正整数的除法

<br />题目描述:编程实现两个正整数的除法,当然不能用除法操作符。<br />// return x/y.<br />int div(const int x, const int y) <br />{<br />          ....<br />}<br />别小看这道题目啊,这个是MS的面试题目啊。刚拿到这道题目的时候。我想过用二分法来求,但二分也还不是要用到除法么。后来想到了一种方法,这种方法我觉得还是可行的,就是把正数转换为字符串,模拟小学学过的竖式除法运算。这样就很快能出结果了,但有一点不

Coding For Children! 7945

轨迹优化 | 图解欧氏距离场与梯度场算法(附ROS C++/Python实现)

距离场在运动规划中则常用于地图建模和轨迹梯度优化。本文图解欧氏距离场EDT的算法原理,并基于双线性插值给出欧氏梯度场的公式推导,并给出ROS C++和Python的算法实现加深理解

FRIGIDWINTER的博客 4648

公开 学生课堂行为数据集 SCB-Dataset: A Dataset for Detecting Student and Teacher Classroom Behavior

公开 学生课堂行为数据集 SCB-Dataset Student Classroom Behavior dataset

CSPhD-winston的博客 1万+

不使用除法来计算两个正整数的除法操作

昨天表哥考我一道算法题,说是在面试中遇到的:如何不使用除法操作符来计算两个正整数的除法操作 太长时间没刷题脑回路跟不上,回来想了想其实很简单,利用二进制的原理,不断求除数的2的幂次方,从而快速逼近最大商 代码很简单,一看就明白了 #include<iostream> using namespace std; void divide(int n,int m) { int s=...

Xylon的博客 776

【TC397第2篇】TC397 EB MCAL开发指南:ADS MCAL Demo集成深度解析

通过本文的指导,我们完成了TC397 EB MCAL与ADS的深度集成,并成功运行了一个简单的MCAL Demo。这个案例不仅帮助我们理解了MCAL在实际开发中的应用,同时也展示了ADS作为一个强大的开发工具在嵌入式系统开发中的优势。在接下来的系列文章中,我们将进一步探讨MCAL的高级应用、优化技巧以及更多实用的开发案例。敬请关注,共同探索TC397 EB MCAL开发的更多精彩!

一直在水些技术小文 452

微软面试题:编程实现两个正整数的除法

编程实现两个正整数的除法,当然不能用除法操作符。 // return x/y. int div(const int x, const int y) { .... } // return x/y int div(const int x, const int y) { int left_num = x; int result = 0; whi...

weixin_33989058的博客 175

win10 跳过U8环境检测的IE WebControls.rar

win10 安装u8的时候 提示IE WebControls 不通过,这个工具可以欺骗u8的检查。 跳过U8环境检测的IE WebControls。提示导入失败后,重新检测通过。

STM32F407中文手册

STM32f407中文手册 本参考手册面向应用开发人员,提供有关使用 STM32F405xx/07xx、STM32F415xx/17xx、 STM32F42xxx 和 STM32F43xxx 微控制器存储器与外设的完整信息。STM32F405xx/07xx、STM32F415xx/17xx、STM32F42xxx 和 STM32F43xxx 构成一个微 控制器系列,各产品具有不同的存储器大小、封装和外设。有关订购信息以及器件的机械与电气特性,请参见数据手册。有关 ARM Cortex™-M4F 内核的信息,请参见《Cortex™-M4F 技术参考手册》。

位运算之美——用+,-和位运算实现正整数除法和取模(一)

9月21日,对本文从格式到部分内容上都进行了修改 另外,鉴于某些转载没有注明出处,考虑到版权问题,特声明如下: 作者:翼帆@cppblog 原文地址:http://www.cppblog.com/xiaoyisnail/archive/2009/09/19/96707.

zhongjie的专栏 2844

文档:亲测可用解决vnc灰屏,黑屏,鼠标大黑叉情况

主要解决vnc-viewer灰屏,黑屏,鼠标大黑叉等现象,详细见文档。

C++程序分享--常见算法/编程面试题:给定一个整数,用辗转相除法实现求解最大公约数

辗转相除法,也被称为欧几里得算法,是一种用于计算两个非负整数a和b的最大公约数(GCD)的算法。这个算法的基本原理是:两个整数的最大公约数等于其中较小的数和两数相除余数的最大公约数。在算法的执行过程中,辗转相除法首先用较大的数除以较小的数,计算出余数。然后用这个较小的数除以余数,再次计算出新的余数。接着,再用这个新的余数除以再次出现的余数,如此反复,直到余数为0为止。有意找工作的同学,请参考博主的原创:《

借雨醉东风的博客 409

最新的sdio 4.0 spec 协议标准 带书签 高清完整PDF版

Part-E1-SDIO-Specification-Ver4.00-Final-120220.pdf:最新的sdio 4.0 spec 协议标准 带书签 高清完整PDF版,包括SDIO 接口的外设,SDIO 接口官方协议等。

106@365大型互联网企业面试题整理&…

全新整理:微软、谷歌、百度等公司经典面试 100 题[第 101-170 题] 整理:July、二零一一年三月九日。 应网友承诺与要求,全新整理。转载,请注明出处。 博主说明: 此 100 题 V0.2 版,本人不再保证,还会提供答案。 因为之前整理的微软 100 题,已经基本上,把题目都出尽了。见谅。 --------------------------------------------微软十

IT小生 2057

《剑指 Offer》专项突破版 - 面试题 1 : 整数除法

详解面试题 1 : 整数除法

melonyzzZ的博客 1233

微软、谷歌、百度等公司经典面试题

全新整理:微软、谷歌、百度等公司经典面试100题[第101-160题] 整理:July、二零一一年三月九日。 应网友承诺与要求,全新整理。转载,请注明出处。 博主说明: 此100题V0.2版,本人不再保证,还会提供答案。 因为之前整理的微软100题,已经基本上,把题目都出尽了。见谅。 --------------------------------------------

Have?Fish:Bear's paw 4657

微软笔试面试题集锦

(一)编程实现两个正整数的除法 编程实现两个正整数的除法,当然不能用除法操作符  想法是这样的:被除数不断减去含除数的最大2^n倍数。例如,200/5,那么以此为200-5*2^5=40 result = 2^5 =32 ,40-5*2^3=0 result=32+2^3 = 40。所以结果为40. int div(const int x, const int y) { int

Solidfish的专栏 778

面试题系列(1)

http://blog.youkuaiyun.com/v_JULY_v/article/details/6234496 微软十五道面试题 1、有一个整数数组,请求出两两之差绝对值最小的值, 记住,只要得出最小值即可,不需要求出是哪两个数。 2、写一个函数,检查字符是否是整数,

628
上一篇: @陈利人 : #面试题#给定两个数组X和Y,元素都是正数。请找出满足如下条件的数对的数目: x^y > y^x,即x的y次方>y的x次方;x来自X数组,y来自Y数组
下一篇: @陈利人 : #面试题#LIS问题:这个LIS问题,可不是Longest Increasing Subsequence,而是Largest Independent Set。
p11323
博客等级 码龄14年 79粉丝 197原创
评论 2
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付:10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值