My father asked __ to help with his work.A.I and TomB.Tom and meC.me and TomD.Tom and I

第1题

Using "I" or "My' instead of "most people" or "our group" ______ message credibility.A decreasesB increasesC has no effect onD distortsE A and B

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第2题

A: Would you like to come to ( )shop one day? B: Yes, I'd love to. Thank you.A. myselfB. meC. my

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第3题

I, my, will, true, come, hope, dream__________________________________________________________________________.

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第4题

阅读下列函数说明和C函数，将应填入(n)处的字句写在对应栏内。[说明]Kruskal算法是一种构造图的最小生成树的方法。设G为一无向连通图，令T是由G的顶点构成的于图，Kmskal算法的基本思想是为T添加适当的边使之成为最小生成树：初始时，T中的点互相不连通；考察G的边集E中的每条边，若它的两个顶点在T中不连通，则将此边添加到T中，同时合并其两顶点所在的连通分量，如此下去，当添加了n-1条边时，T的连通分量个数为1，T便是G的一棵最小生成树。下面的函数void Kruskal(EdgeType edges[],int n)利用Kruskal算法，构造了有n个顶点的图 edges的最小生成树。其中数组father[]用于记录T中顶点的连通性质：其初值为father[i]=-1 (i=0,1,…,n-1)，表示各个顶点在不同的连通分量上；若有father[i]=j，j＞-1，则顶点i，j连通；函数int Find(int father[],int v)用于返回顶点v所在树形连通分支的根结点。[函数]define MAXEDGE 1000typedef struct{ int v1;int v2;}EdgeType;void Kruskal(EdgeType edges[],int n){ int father[MAXEDGE];int i,j,vf1,vt2;for(i=0;i＜n;i+ +) father[i]=-1;i=0;j=0;while(i＜MAXEDGE && j＜(1)){ vf1=Find(father,edges[i].v1);vf2=Find(father,edges[i].v2);if((2)){(3)=vf1;(4);printf("%3d%3d\n",edges[i].v1,edges[i].v2);}(5);}}int Find(int father[],int v){ int t;t=v;while(father[t]＞=0) t=father[t];return(t);}

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第5题

I( )a cup of coffee, but they gave me a cup of tea.A.asked forB.asked

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第6题

I'd love to()with you my new kitchenware.A. shareB. putC. help

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第7题

下列程序的输出结果是【 】。 includeclass MyClass{public: int number; void set(in下列程序的输出结果是【 】。include＜iostream. h＞class MyClass{public:int number;void set(int i);};int number=3;void MyClass: :set (int i){number=i;}void main( ){MyClass my1；int number=10;my1.set(5),cout＜＜my1, number＜＜',';my1.set(number);cout＜＜my1.number＜＜',';my1.set(: :number);cout＜＜my1.number＜＜'.';}

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