1 for x in range(0,100): 2 for y in range(0,100): 3 for z in range(0, 100): 4 if x*3+y*2+0.5*z == 100 and x+y+z==100: 5 print(x,y,z)

LeetCode面试常见100题( TOP 100 Liked Questions) 置顶 2018年07月16日 11:25:22 lanyu_01 阅读数 9704更多 分类专栏: 面试编程题真题合集 常见算法问题 LeetCode试题 LeetCode常见试题   版权声明:本文为博主原创文章,遵循CC 4.0 BY-SA版权协议,转载请附上原文出处链接和本声明. 本文链接:https://blog.csdn.net/lanyu_01/article/details/81062232 这篇

桌面上运行virtualbox出错: The virtual machine 'xp' has terminated unexpectedly during startup with exit code 1 (0x1). Effective UID is not root(euid=1000 egid=482 uid=1000 gid=482)(rc=-10) Please try reinstalling VirtualBox. 手动敲命令运行/usr/lib/virtual/VirtualB

题目链接:https://vjudge.net/problem/POJ-2299 推荐讲解树状数组的博客:https://blog.csdn.net/int64ago/article/details/7429868 题目意思就是让我们把无序的一些数字经过相邻数字间两两交换,最后变成不递减的数字.我们要求出最少要交换的次数. 逆序对(百度的):对于一个包含N个非负整数的数组,A[1..n],A[1..n],如果有i < j,且A[ i ]>A[ j ],则称(A[ i] ,A[ j] )为数组A

例11  求质数 问题描述 质数是指除了有1和自身作为约数外,不再有其他约数的数.比如:3.5.7是质数.而9不是质数,因为它还有约数3. 编写程序求给定区间中的所有质数. 输入格式 两个整数a和b,其中1≤a≤b≤100000. 输出格式 输出给定范围的所有质数,输出时每个质数占5列,每行输出10个质数. 输入样例 100  200 输出样例 101  103  107  109  113  127  131  137  139  149 151  157  163  167  173  17

文件最后100行:tail -n100 filePath: 文件开头100行:head -n100 filePath: 文件指定开始行和结束行的内容:sed '1,100p' filePath: 文件有多少行:wc -l 文件名:

<div style="width:100px; background-color: aqua; min-width:100%">kkk</div>

3 down vote Just run the command java -X and you will get ans of all_ C:\Users\Admin>java -X -Xmixed mixed mode execution (default)-Xint interpreted mode execution only -Xbootclasspath:<directories and zip/jar files separated by ;> set search pat

create table t( id ,), name ),--商品名称 j int, --入库数量 c int, --出库数量 jdate datetime --入库时间 ) ,,'2007-12-01' ,,'2008-01-07' ,,'2007-12-21' ,,'2008-01-15' ,,'2008-02-03' ,,'2008-02-01' ,,'2008-03-01' go create proc wsp ),--商品名称 @cost int --销售量 as --先得出该货物的

题目地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2588 Burning Bridges Time Limit: 5 Seconds      Memory Limit: 32768 KB Ferry Kingdom is a nice little country located on N islands that are connected by M bridges. All bridges are very

Ponds Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1288    Accepted Submission(s): 429 Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, a

1003 Emergency (25分)   As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between

Primitive Roots Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3381   Accepted: 1980 Description We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is eq

PS:自己写的,自测试OK,供大家参考. /* 一条长廊里依次装有n(1 ≤ n ≤ 65535)盏电灯,从头到尾编号1.2.3.…n-1.n.每盏电灯由一个拉线开关控制.开始,电灯全部关着.有n个学生从长廊穿过.第一个学生把号码凡是1的倍数的电灯的开关拉一下:接着第二个学生把号码凡是2的倍数的电灯的开关拉一下:接着第三个学生把号码凡是3的倍数的电灯的开关拉一下:如此继续下去,最后第n个学生把号码凡是n的倍数的电灯的开关拉一下.n个学生按此规定走完后,长廊里电灯有几盏亮着.注:电灯数和学生数一致

j package test1; //2018/11/30 //求100以内的所有素数 public class Main10 { public static void main(String[] args){ Main10 test=new Main10(); System.out.println("100以内的素数:"); int count=0; for(int i=2;i<100;i++) { if(test.isPrimeNumber(i)==true) { Syste

def get_height(n): if n==1: eturn 150 return 100+sum([200*pow(0.5,i) for i in range(1,n)])+100*pow(0.5,n) for i in range(1,5): print get_height(i) 150 225.0 262.5 281.25结果如上, 递归法求解如下:def get_height(n): if n==1: eturn 150 return get_height(n-1)+100*po

需求说明: 用100文钱买鸡,公鸡5文钱一只,母鸡3文钱一只,小鸡3只1文钱,要求公鸡.母鸡.小鸡都必须要有,刚好用完100文钱,公鸡.母鸡.小鸡的数量之和也是100. public class test { public static void main(String[] args) { int cock,hen,chicken;//公鸡,母鸡,小鸡 for(cock=1; cock<=20; cock++){ for(hen=1; hen<=33; hen++){ for(chicken=

a = [100]  #每个‘反弹落地’过程经过的路程,第1次只有落地(100米) h = 100  #每个‘反弹落地’过程,反弹的高度,第1次为100米 print('第1次从%s米高落地,走过%s米,之后又反弹至%s米.' % (h, a[0], h/2)) for i in range(2,11):  #第1次已初始化,再循环9次     a.append(h)  #先计算路程,再高度减半,因为一个‘反弹落地’为2个高度     h = h / 2     print('第%s次从%s米高

<计算机网络>谢希仁著第四版课后习题答案答: 数据长度为100字节时 传输效率=100/(100+20+20+18)=63.3% 数据长度为1000字节时, 传输效率=1000/(1000+20+20+18)=94.5%

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