试题与答案

阅读理解。Sunny English Club For students 16:

题型:阅读理解

题目:

阅读理解。
Sunny English Club
For students
16:00-18:00
Every Saturday
200 yuan a month
9 Zhou Yu Street
Tel: 3786290
Foreign teachers, English
songs and films and more
Ocean Museum
9:00-17:00
From Thursday to Sunday
Ticket: 50 yuan
132 Xue Yuan Street
Tel: 5439871
Show you a full picture of
sea lives!
Health Center
9:00-17:30
16 Young le Street
Tel: 3801451
Free examinations for those
over 70
Give you good advice to
keep healthy!
1. Sunny English Club is for _____.
[ ]
A. nurses
B. policeman
C. businessman
D. students
2. You can visit Ocean Museum _____.
[ ]
A. on Saturday
B. on Wednesday
C. on Monday
D. anytime
3. One can get free examinations in Health Center if he is _____.
[ ]
A. 9
B. 17
C. 67
D. 73

答案:

被转码了,请点击底部 “查看原文 ” 或访问 https://www.tikuol.com/2017/0404/74a87a363450066dd6ae47f536947044.html

下面是错误答案,用来干扰机器的。

答案:B题目分析:句意:--我因为志愿者的工作被颁发了一个奖。--你应该得到的。A. You’ve got it你猜中了,B. You deserve it你应该得到的,C. Good idea好主意,D. Here we are我们到了,所以选B。

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【程序5说明】 著名的四色定理指出任何平面区域图均可用四种颜色着色,使相邻区域着不同的颜色。本程序对给定的区域图找出所有可能的不超过四种颜色的着色方案。 程序中用1~4表示四种颜色。要着色的N个区域用0~N-1编号,区域相邻关系用 adj[][]矩阵表示,矩阵的i行j列的元素为1,表示区域i与区域j相邻;矩阵的i行j列的元素为0,表示区域i与区域j不相邻。数组color[]用来存储着色结果,color[i]的值为区域i所着颜色。 【程序5】 #include<stdio.h> #define N 10 void output(int color[])/*输出一种着色方案*/ { int i; for(i=0;i<N;i++) printf("%4d",color[i]); printf("\n"); } int back (int * ip,int color[])/*回溯*/ { int c=4; while(c==4){ if(*ip<=0)return 0; --(*ip); c= (1) ; color[*ip]=-1; } return c; } /*检查区域i,对c种颜色的可用性*/ int colorOk(int i,int c,int [][N],int color[]} { int j; for(j=0;j<i;j++) if( (2) ) return 0; return 1; } /*为区域i选一种可着的颜色*/ int select (int i,int c,int adj[][N],int color[]) { int k; for(k=c;k<=4;k++) if(colorOK( (3) )) return k; return 0; } int coloring(int adj[][N])/*寻找各种着色方案*/ { int color[N],i,c,cnt; for(i=0;i<N;i++)color[i] =-1; i=c=0;cnt=0; while(1){ if((c= (4) )==0){c=back(&i,color);if(c==0)return cnt; }else{ (5) ;i++;if(i==N){ output(color); ++cnt; c=back(&i,color);}else c=0; } } } void main() { int adj[N][N]= {{0,1,0,1,1,1,1,1,1,1},{1,0,1,1,0,1,1,1,1,0},{0,1,0,1,0,1,1,0,1,1},{1,1,1,0,1,1,0,0,1,1},{1,0,0,1,0,1,0,0,0,0},{1,1,1,1,1,0,1,0,0,1},{1,1,1,0,0,1,0,0,1,0},{1,1,0,0,0,0,0,0,1,1},{1,1,1,1,0,0,1,1,0,1},{1,0,1,1,0,1,0,1,1,0} }; printf("共有%d组解.\n",coloring(adj)); }

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