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何谓斯笃克定律(Stoke's
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内容提示：(1)μ=水之粘滞系数 (2)γs=土粒单位重 (3)Zr'=比重计读数至比重计浮力中心之距离 2-16 比重计分析试验需用到下列公式,各参数之意义何者有误:
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何谓斯笃克定律(Stoke's
官方公共微信What's color is the iacket?句子中的the为什么不改成a?
如果用a,就是泛指夹克这种物品,你是想问夹克都可以是什么颜色么?这句话是问那件夹克是什么颜色,应该有前后语境的.
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扫描下载二维码I'll join in the teacher's relay race为什么要用join in
小丑鱼jzAO
join指加入某党派、某组织或某社会团体,以及参军等,例句：①I will never forget the day when I joined the Party.我永远也忘不了入党的那一天.②Will you join us for dinner?请你和我们一起吃饭好吗?join in通常指参加某种活动（注意是活动）,尤其指和其他人一起参加某项活动.常与take part in进行转换.例句：①May I join in the game?我可以参加这个比赛吗?②I hope you”ll all join in the discussion.我希望你们大家都参加讨论.【补充说明!】：join in后面加的活动通常指正在进行的.这句话的意思是 我参加教师接力赛.（是参加某种活动,且是和其他人一起参加的,所以用 in）反之,若不强调这个MOVEMENT正在进行,就不用加in.
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扫描下载二维码POJ 1634 Who's the boss?
POJ 1634 Who's the boss?，有需要的朋友可以参考下。Who's the boss?Time Limit:5000MSMemory Limit:30000KTotal Submissions:2773Accepted:1142DescriptionSeveral surveys indicate that the taller you are, the higher you can climb the corporate ladder. At TALL Enterprises Inc. this &de facto standard& has been properly formalized: your boss is always at least as tall as you are. Furthermore, you can safely assume that your boss earns a bit more than you do. In fact, you can be absolutely sure that your immediate boss is the person who earns the least among all the employees that earn more than you and are at least as tall as you are. Furthermore, if you are the immediate boss of someone, that person is your subordinate, and all his subordinates are your subordinates as well. If you are nobody's boss, then you have no subordinates. As simple as these rules are, many people working for TALL are unsure of to whom they should be turning in their weekly progress report and how many subordinates they have. Write a program that will help in determining for any employee who the immediate boss of that employee is and how many subordinates they have. Quality Assurance at TALL have devised a series of tests to ensure that your program is correct. These test are described below.InputOn the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each scenario begins with a line containing two positive integers m and q, where m (at most 30000) is the number of employees and q (at most 200) is the number of queries. The following m lines each list an employee by three integers on the same line: employee ID number (six decimal digits, the first one of which is not zero), yearly salary in Euros and finally height in μm (1 μm = 10-6meters - accuracy is important at TALL). The chairperson is the employee that earns more than anyone else and is also the tallest person in the company. Then there are q lines listing queries. Each query is a single legal employee ID.The salary is a positive integer which is at most 10 000 000. No two employees have the same ID,and no two employees have the same salary. The height of an employee is at least 1 000 000 μm and at most 2 500 000 μm.OutputFor each employee ID x in a query output a single line with two integers y k, separated by one space character, where y is the ID of x's boss, and k is the number of subordinates of x. If the query is the ID of the chairperson, then you should output 0 as the ID of his or her boss (since the chairperson has no immediate boss except, possibly, God).Sample Input23 23 8 21 74 34 3 45 1 1Sample Output
3SourceNorthwestern Europe 2003ACcode：#include &map&#include &queue&#include &cmath&#include &cstdio&#include &cstring&#include &stdlib.h&#include &iostream&#include &algorithm&#define maxn 30010int hash[1000000];int stack[maxn],top=0;struct woker{
void in(){scanf(&%d%d%d&,&id,&wage,&height);boss=0;nsub=1;
void on(){printf(&id %d wage %d height %d boss %d nsub %d/n&,id,wage,height,boss,nsub);
}}my[maxn];bool cmp(woker a,woker b){
return a.wage&b.}int main(){
int loop,n,m;
scanf(&%d&,&loop);
while(loop--){scanf(&%d%d&,&n,&m);for(int i=0;i&n;++i)my[i].in();sort(my,my+n,cmp);for(int i=0;i&n;++i){hash[my[i].id]=i;}top=0;for(int i=0;i&n;i++){while(top&&my[stack[top-1]].height&=my[i].height){
my[stack[top-1]].boss=my[i].
my[i].nsub+=my[stack[top-1]].
stack[top++]=i;}/**for(int i=0;i&n;++i){printf(&%d&,i);my[i].on();}**/int t,k;for(int i=0;i&m;++i){scanf(&%d&,&t);k=hash[t];printf(&%d %d/n&,my[k].boss,my[k].nsub-1);}
return 0;} 版权声明：本文为博主原创文章，未经博主允许不得转载。
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