西北工业大学

Mother lets me play football _____ I've finished homework.A.unlessB.untilC.only ifD.if only

题目
Mother lets me play football _____ I've finished homework.

A.unless

B.until

C.only if

D.if only

参考答案和解析
正确答案:C
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相似问题和答案

第1题:

I_____ jogging every morning in the park.

A. go

B. do

C. play


参考答案:A

第2题:

I _________ table tennis quite well, but I haven ’t had time to play since the New Year.

A. will play

B. have played

C. play


参考答案:C

第3题:

A wise mother never ______ her children to the slightest possibility of danger.

(A) exposes

(B) makes

(C) sends

(D) lets


正确答案:A
解答参考:A.本题考查expose的用法。expose somebody to表示使某人接触……,使处于……影响下‖。【译文】明智的母亲绝不会让孩子们处于哪怕是一丁点的可能的危险当中。

第4题:

根据下列程序的执行结果,可以断定划线部分的修饰符应为______。执行结果:

i9a football is created.

a football is created.

i=10

a football is destroyed.

a football is created.

i=20

a football is destroyed.

a football is destroyed.

源程序:

include<iostream.h>

class Football

{

public:

Football( ){cout<<"a football is created."<<endl;}

~Football( ){tout<<"a football is destroyed."<<endl;}

};

void func(int i)

{

______Football f1;

Football f2;

cout<<"i="<<i<<endl;

}

void main( )

{

func(10);

func(20);

}


正确答案:static
static 解析:观察一下输出中对象的构造函数和析构函数的执行次数可以看出,再次进入时某对象没有被构造,亦即该对象在func执行后没有被销毁,再次进入时则不需要重构,可以体现这种功能的标识符只有static。

第5题:

–– Jack: Mommy, when will we have dinner? I'm starving.

–– Mother:————————? We have to wait for daddy.

A: Quickly, honey

B: OK, honey

C: Soon, honey

D: All right, honey


参考答案:C

第6题:

After they finished _____football,they went for a drink in a pub.

A.playing

B.to play

C.play


参考答案:A

第7题:

I've always enjoyed_____ football.

A. to play

B. playing

C. play


参考答案:B

第8题:

It is ______ who ______ wrong.

A.me; me

B.me; has

C.I; am

D.I; is


参考答案:C

第9题:

I() ping-pang quite well, but I haven t had time to play since the new year.

A、will play

B、have palyed

C、played

D、play


参考答案:D

第10题:

B 宽度优先(种子染色法)

5.关键路径

几个定义: 顶点1为源点,n为汇点。

a. 顶点事件最早发生时间Ve[j], Ve [j] = max{ Ve [j] + w[I,j] },其中Ve (1) = 0;

b. 顶点事件最晚发生时间 Vl[j], Vl [j] = min{ Vl[j] – w[I,j] },其中 Vl(n) = Ve(n);

c. 边活动最早开始时间 Ee[I], 若边I由<j,k>表示,则Ee[I] = Ve[j];

d. 边活动最晚开始时间 El[I], 若边I由<j,k>表示,则El[I] = Vl[k] – w[j,k];

若 Ee[j] = El[j] ,则活动j为关键活动,由关键活动组成的路径为关键路径。

求解方法:

a. 从源点起topsort,判断是否有回路并计算Ve;


正确答案: