万维考试系统-C语言题库-程序设计题导出-113道.doc

King shop school car to watch the summary view store is located in the LAN-Xin railway, 215 students, and students to and from school has about 90 people to cross the railway, more than half of the students live in the school in the vicinity of the rail line along the railway more than 10 square kilometers. Railway is economy of artery, to makes railway along general people master railway legal regulations and security knowledge, enhanced love road watch consciousness, ensure railway security smooth, years, I school has been put love road watch work as education students of focus to caught, made has significantly of effect, years no occurred had railway security accident, emerged out a and a love road watch small guard, General teachers and students and masses of love road watch consciousness General improve, times by superior recognition, Our practices are reporting on is as follows. And improve the Organization, love the way of protection responsibilities. Set up by the principal of the school is the head security administrator for the Deputy Head of the class teacher for members of the railway protection leading group for propaganda and education, consisting of offices, advocacy groups, education groups and steering groups, organizations responsible for railway protection publicity and education of the whole school, command and coordination. To ensure safety of the railway for the overall objectives, major tasks are: 1, arouse the enthusiasm of teachers, school advantages into full play, watch the propaganda and education work of joint railway. 2, railway watch teacher and student awareness, to ensure against road-related casualties. 3, education in primary and middle school students the courage to fight against the Railway Safety Act and found to have damaged railway safety relevant parts of the report in a timely manner to ensure railway safety. Do Division, responsibility and student teacher assessment, evaluation and love road linked to the protection and safety, enhance the sense of responsibility. Second, programming, watch love road work routine. School developed has railway watch publicity education work programme, put love road watch work into school education teaching plans, full using class teaching, and theme class, timing, education students learning railway method, about love road security matters, school requirements daily of campus broadcast to has five minutes of railway security knowledge education, all class weekly are to Shang a section love road watch aspects of security class, monthly introduced a period love road watch aspects of window publicity, reached has alarm ringing of effect. Third, explicitly demanded that the love road protection work. Schools to establish and improve a love road watch schedule, enacted the railway safety ten no Protocol, the Convention on the safety and other regulations and requirements, so as to enable departments to further their work. Which publicity group is responsible for railway watch legal regulations and about common sense of publicity, annual organization a to two times above railway method, and security management Punishment Ordinance, for main content of publicity, further strengthened General teachers and students of concept and love road watch consciousness, education students do three not: that not to railway Shang play, without stones, hit train, is not rail Shang stacked roadblocks. Education groups to include railway protection knowledge education in carrying out quality education, class of railway transport in the main channel of textbooks in economic construction in the role and significance of guiding students to take care of the railway一、程序设计 共113题 （共计1130分）第1题 （10.0分） 题号:390/*-【程序设计】-功能：编写函数求1100中奇数的平方和。 结果为166650.000000。-*/#include void wwjt(); float sum(int n) /*Program*/ /* End */答案：-float s=0; int i; for(i=1;i=n;i=i+2) s=s+i*i; return(s);-第2题 （10.0分） 题号:320/*-【程序设计】-功能：在键盘上输入一个3行3列矩阵的各个元素的值（值 为整数），然后输出主对角线元素的积,并在fun() 函数中输出。-*/#include void wwjt(); main() int i,j,s,a33; for(i=0;i3;i+) for(j=0;j3;j+) scanf(%d,&aij); s=fun(a); printf(Sum=%dn,s); wwjt();int fun(int a33) /*Program*/ /* End */ 答案：-int sum;int i,j;sum=1;for(i=0;i3;i+) sum=sum*aii; return sum;-第3题 （10.0分） 题号:324/*-【程序设计】-功能：能计算从1开始到n的自然数中偶数的平方的和，n由 键盘输入，并在main()函数中输出。（n是偶数）-*/#include void wwjt(); int fun(int n) /*Program*/ /* End */ main() int m; printf(Enter m: ); scanf(%d, &m); printf(nThe result is %dn, fun(m); wwjt(); 答案：-int sum,i;sum =0;for(i=2;i=n;i=i+2)sum=sum+i*i;return(sum);-第4题 （10.0分） 题号:345/*-【程序设计】-功能：将两个两位数的正整数a、b合并形成一个整数放在c 中。合并的方式是：将a数的十位和个位数依次放在 c数的千位和十位上， b数的十位和个位数依次放在 c数的个位和百位上。 例如：当a45，b=12。调用该函数后，c=4251。 -*/#include void wwjt(); void fun(int a, int b, long *c) /*Program*/ /* End */main() int a,b; long c; printf(input a, b:); scanf(%d%d, &a, &b); fun(a, b, &c); printf(The result is: %ldn, c); wwjt(); 答案：-*c=a/10*1000+a%10*10+b/10+b%10*100;- 第5题 （10.0分） 题号:366/*-【程序设计】功能：求一组数中大于平均值的数的个数。例如：给定的一组数为1,3,6,9,4,23,35,67,12,88时，函 数值为3。-*/ #include void wwjt(); int fun(int a,int n) /*Program*/ /* End */ main() int a10=1,3,6,9,4,23,35,67,12,88; int y; y=fun(a,10); printf(y=%dn,y); wwjt();答案：-int i,k=0; float s=0,ave; for(i=0;in;i+) s+=ai; ave=s/n; printf(%f ,ave); for(i=0;iave)k+; return k;-第6题 （10.0分） 题号:355 /*-【程序设计】-功能：对长度为8个字符的字符串，将8个字符按降序排列。例如：原来的字符串为CEAedcab,排序后输出为edcbaECA。-*/#include#include#includevoid wwjt(); void fun(char *s,int num) /*Program*/ /* End */main() char s10; printf(输入8个字符的字符串:); gets(s); fun(s,8); printf(n%s,s); wwjt();答案：- int i,j; char t; for(i=0;inum;i+) for(j=i+1;jnum;j+) if(sisj) t=si;si=sj;sj=t;-第7题 （10.0分） 题号:44/*-【程序设计】-功能：求出二维数组周边元素之和，作为函数值返回。二 维数组的值在主函数中赋予。-*/#define M 4#define N 5#include void wwjt(); int fun(int aMN)/*Program*/* End */main()int aMN=1,3,5,7,9,2,4,6,8,10,2,3,4,5,6,4,5,6,7,8;int y;y=fun(a);printf(s=%dn,y);wwjt();答案：-int s=0; int i,j; for(i=0;iM;i+) s=s+ai0+aiN-1; for(j=1;jN-1;j+) s=s+a0j+aM-1j; return s;-第8题 （10.0分） 题号:339/*-【程序设计】-功能：判断整数x是否是同构数。若是同构数，函数返回1； 否则返回0。x的值由主函数从键盘读入，要求不大 于100。说明：所谓“同构数”是指这样的数，这个数出现在它的 平方数的右边。例如：输入整数5，5的平方数是25，5是25中右侧的数，所 以5是同构数。-*/#include void wwjt(); int fun(int x) /*Program*/ /* End */main() int x,y; printf(nPlease enter a integer numbers:); scanf(%d,&x); if(x100) printf(data error!n); exit(0); y=fun(x); if(y) printf(%d YESn,x); else printf(%d NOn,x); wwjt(); 答案：- int k; k=x*x; if(k%10=x)|(k%100=x) return 1; else return 0;-第9题 （10.0分） 题号:383/*-【程序设计】-功能：用函数实现字符串的复制, 不允许用strcpy()函数。-*/#include void wwjt(); void copy(char str1,char str2) /*Program*/ /* End */main() void copy(); char c140,c240; gets(c1); copy(c1,c2); puts(c2); wwjt();答案：-int i; for(i=0;str1i!=0;i+) str2i=str1i; str2i=0;-第10题 （10.0分） 题号:310/*-【程序设计】-功能：计算并输出给定整数n的所有因子之和（不包括1与 自身）。注意：n的值不大于1000。例如：n的值为855时，应输出704。-*/#include void wwjt(); int fun(int n) /*Program*/ /* End */ main() printf(s=%dn,fun(855); wwjt();答案：-int s=0,i; for(i=2;in;i+) if(n%i=0)s=s+i; return s;-第11题 （10.0分） 题号:354/*-【程序设计】-功能：求小于lim的所有素数并放在aa数组中，该函数返回 所求出素数的个数。-*/#include#include#define MAX 100void wwjt(); int fun(int lim,int aaMAX) /*Program*/ /* End */main() int limit,i,sum; int aaMAX; printf(Please input ainteger:); scanf(%d,&limit); sum=fun(limit,aa); for(i=0;isum;i+) if(i%10=0&i!=0) printf(n); printf(%5d,aai); wwjt();答案：- int n=0; int i,j; for(i=2;i=lim;i+) for(j=2;ji;j+) if(i%j=0) break; if(j=i) aan+=i; return n;-第12题 （10.0分） 题号:341/*-【程序设计】-功能：删除所有值为y的元素。数组元素中的值和y的值由 主函数通过键盘输入。-*/#include #include#include#define M 20void wwjt(); void fun(int bb,int *n,int y) /*Program*/ /* End */main() int aaM,n,y,k; printf(nPlease enter n:);scanf(%d,&n); printf(nEnter %d positive number:n,n); for(k=0;kn;k+) scanf(%d,&aak); printf(The original data is:n); for(k=0;kn;k+) printf(%5d,aak); printf(nEnter a number to deletede:);scanf(%d,&y); fun(aa,&n,y); printf(The data after deleted %d:n,y); for(k=0;kn;k+) printf(%4d,aak); printf(n); wwjt();答案：- int i,j; for(i=0;i*n;) if(bbi=y) for(j=i;j*n;j+) bbj=bbj+1; *n=*n-1; else i+; -第13题 （10.0分） 题号:317/*-【程序设计】-功能：求出NM整型数组的最大元素及其所在的行坐标及 列坐标（如果最大元素不唯一，选择位置在最前面 的一个）。例如：输入的数组为: 1 2 3 4 15 6 12 18 9 10 11 2 求出的最大数为18,行坐标为2，列坐标为1。-*/#define N 4#define M 3#include void wwjt(); int Row,Col;int fun(int arrayNM) /*Program*/ /* End */ main() int aNM,i,j,max; printf(input a array:); for(i=0;iN;i+) for(j=0;jM;j+) scanf(%d,&aij); for(i=0;iN;i+) for(j=0;jM;j+) printf(%d,aij); printf(n); max=fun(a); printf(max=%d,row=%d,col=%d,max,Row,Col); wwjt();答案：-int max,i,j;max=array 00;Row=0;Col=0;for(i=0;iN;i+) for(j=0;jM;j+) if(maxarray ij) max=array ij; Row=i;Col=j;return(max);-第14题 （10.0分） 题号:392/*-【程序设计】-功能：求一批数中最大值和最小值的积。-*/#define N 30#include stdlib.h#include void wwjt(); int max_min(int a,int n) /*Program*/ /* End */main() int aN,i,k; for(i=0;iN;i+) ai=random(51)+10; for(i=0;iN;i+) printf(%5d,ai); if(i+1)%5=0) printf(n); k=max_min(a,N); printf(the result is:%dn,k); wwjt();答案：- int i,max,min; max=min=a0; for(i=1;imax) max=ai; else if(aimin) min=ai; return(max*min);-第15题 （10.0分） 题号:501#include /*-【程序设计】-题目：从键盘输入一个大写字母，要求改用小写字母输出。-*/#include void wwjt(); char fun(char c) /*Program*/ /* End */ void main() char c1, c2; printf(Please input a char:); c1=getchar(); printf(%c,%dn,c1,c1); c2=fun(c1); printf(%c,%dn,c2,c2); wwjt(); fclose(IN); fclose(OUT);答案： c=c+32; return c;第16题 （10.0分） 题号:502题目：用while语句求1100的累计和。#include /*-void wwjt(); int fun(int n) /*Program*/ /* End */ void main() int sum = 0; sum=fun(100); printf (sum = %dn, sum); wwjt();答案： int i=1,sum=0; while(i=n) sum=sum+i; i+; return sum;第17题 （10.0分） 题号:409/*-【程序设计】-题目：输入华氏温度求摄氏温度。转换公式为 c=5/9（f-32）， 输出结果取两位小数。-*/#include void wwjt(); double fun(double m) /*Program*/ /* End */ void main() double c,f; printf(请输入一个华氏温度:); scanf(%f,&f); c=fun(f); printf(摄氏温度为:%5.2fn,c); wwjt();答案： float n; n=(5.0/9.0)*(m-32); return n; 第18题 （10.0分） 题号:29/*-【程序设计】-功能：对任意输入的 x，用下式计算并输出 y 的值。 5 x10-*/#includevoid wwjt();int fun(int n) /*Program*/ /* End */ void main () int x,y; printf(enter x:); scanf(%d,&x); y=fun(x); printf(x=%d,y=%dn,x,y); wwjt();答案： int m; if(n=10) if(n10) m=-5; else m=0; else m=5;return m;第19题 （10.0分） 题号:382/*-【程序设计】-功能：给定n个数据, 求最小值出现的位置（如果最小值 出现多次,求出第一次出现的位置即可）。-*/#include void wwjt(); int station(int s,int n) /*Program*/ /* End */main() int a100,n,i,t; scanf(%d,&n); for(i=0;in;i+) scanf(%d,&ai); t=station(a,n); printf(th