湖南工业大学C语言上机操作考试题(1)

1、C语言机试模拟试题一、改错题：(30分)1、求一维数组a中的值为偶数的元素之和。#include <stdio.h>siun (iiit aii ,iiit n) iiit i,s;s = 0;for (i=0; i<n; i-H-)if (aiii % 2 = 0)/千不帀于于于乔于下于千TO1 111(i于于不下下于千千不不于千/s=s+i;改为：s = s + arri;retiun (s);void maiii() iiit a10=10,4,2,7,3,12,5,34,5,9,i,s;/平平平平平千平于平平平平Tcm、ci于平千千平于千平于千平于/s = sum(

2、a ,2 );改为：s = sum( a ,10 ); printf("The result is: %dn", s);2、求一维数组a中的值为奇数的元素之和。#iiiclude <coiiio.h>卄include <stdio.h>iiit sum( iiit b ,iiit n) iiit i,s = 0;for (i=0; i<n; i-H-)if (bi % 2 = 1)y* * */s = s + bi改为：retiun (s);void inaiii() iiit a12=10,4,2,7,3,12,5,34,5,9,21,19,

3、ii;9Y/下不千于于节下于节千于节re Hi(i于于不不下于千节不不于干/n = sum(a,2);改为： piiiitf("The result is :%dii'n );3、求二维数组a中的最大值和最小值。#iiiclude <stdio.h>void maiii()iiit a33=4,4,34,37,3,12,5,6,5,i,j,max»min;/cliscr();max = niiii = a00;for (i=0; i<3; i卄)>Y/干千平于千千平于千千千千Tcm】Ci于千于下平于千千千于于千/for (j=l; j<

4、3; j-H-)改为:for (j=0; j<3; jF) if(max<aij) max = aij;/平平平平平平平于平于平平tc】m c于于平千平于平千于千平于/ if(niiii<aij)改为：if (min > aiUJ) nwi = aij;printfpThe max is: %dnn, max); printf("The min is: %dn", min);/求一维数组a中的最大元素及其下标。#iiiclude <stdio.h>void maiii()iiit a10=l,4,2,7,3,12,5,34,5,9,i,m

5、ax,pos;/cliscr();max = a0;pos = 0;for (i=l; i<10; i-H-)9Y/平不于于平下下于于于平IO1 111f|于于平平下于于平千于于于/if (max > ai)改为：if (max< ai)max = ai;i = pos;改为：pos = i; piintfpThe max is: %d、pos is: %dir max , pos);5猴子吃桃的问题，求第一天的桃子数。#iiiclude <stdlib.h>#include <math.h>void maiii() iiit i,j,x,p; /cl

6、i'scr(); x=l;j=12;改为：j=ll;wliile(j>=l) p=(x+l)*2;/xlx/平平平于于乎于不于于平下Tc hi门 于平平吓平于平平平方平卒/P=X：改为：x = p;j“；pi'iiitf(ntotal is %dn",p);6、求一维数组a中的值为奇数的元素的平均值。#iiiclude vstdio.h>double average(iiit an , int n) iiit k=0,i; double s;s = 0;for (i=0 ;i<n; i-H-)if (arri % 2 = 1)改为：if (arri

7、% 2 = =1) s = s + aiyi; k-H-; retiun (s/k);void maiii() inta12=10,4,2,7,3,12,5,34,5,9,21,18; double s;/ cliscr();s = average(a12,12);改为:s = average(a, 12); printffThe result is: %.2fn", s);二、填空题：(30分)1、程序的功能是：在第一个循环中从键盘上给a数组的前M (M<100)个数组元素依次赋值，在第二个循环中使a数组前M个元素中的值对称折叠，在最后一个循环中输出折叠后的a数组 的前M个元

8、素。程序如下：#defiiie M 5void maiii() int i,a100,t;/clrscr();for(i=0;i<M;i-H-)(1)填写的语句为：scaiif(-%d-,&ai);/Yf/于下平于于平于下于平平平于于t"C 1T1 f|平平平平于平下平平于平平/for(i=0;_(2)_;i+)-填写的语句为:ivM/2 aM-i-l=ai;for(i=0;i<M;i-H-)printf(n%5drai);printf(”n”)；2、计算正整数num的各位上的数字之和。#iiiclude vstdio.h>void maiii() ( ii

9、it num,k;/cli'scr();(1);填写的语句为：k=0; piiiitf("Please enter a munber:n); scanf(”d",&num);dok= (2) ;填写的语句为：k=k+mun% 10;mmi/=10; wliile(nxun);printf(”n%dn”,k);email is 663、求cmaii=in!/n! *(mn)!之值，例如:m=12,n=2时, #iiiclude <inath.h># include <st±o.h>long iiit fiui(int x)lo

10、ng int s=l; iiit i; for (i=l;iv=x：i+) s*=i;9!/节不乔于于于下于不于于IO1 111 Ci于于不不帝于千不不不于于/(1)；填写的语句为：retimis;void main() long iiit m=12,n=2,sl,s2,s3;/clrscr();sl=fim(m);s2=fiui(n);s3=fiui(m-n);piiiitf(nCnm is : %ld",(2);填写的语句为：sl/(s2*s3) 4、输出Fabonacci数列1, 1, 2, 3, 5前20项，要求每位输出5个 数。#iiiclude "stdio.h

11、" void maiii() longiiit f20=l,l; iiit i;/ckscr();J1 YJi千于于于千千平于节千千于平于TcimC于千平于千千千于平千千平/(1)；填写的语句：intj=O;for(i=2;i<20;i-H-)fi=fil+fli2;JJ/千干节于千千于于节千千于于千TcimC千千于于干千千于于千干千/if(2) piintf("n");填写的语句：j卄5=0piintf(n%81d,',fi);printf(”n”)；5、求两个正整数m,n之间所有既能被3整除也能被7整除的整数之 和。#iiiclude <m

12、ath.h> #mclude <stdio.h>main()(iiit mjiXt; long iiit s=0; /c1isct();scanf("%d,%d”,&m, &n); if( m>n)t=in; m=n; n=t;for (1)；讦+)填写的语句：i=m;i<=n;if( i%3=0 && i%7=0 )s += i;9Y/不帝不于于于下于节于干节tC"!"IT】(i于于不不节于千不不不于于/printf("Sum is :(2)n",s );填写的语句：%ld6、从键

13、盘输入一个由数字组成的字符串(字符串长度小于8),将 该字符串转换成一个十进制数。例如：从键盘输入12345,则程 序运行的结果应当为：n=12345#iiiclude "stdio.h" #iiiclude nstiiiig.hM void maiii() char s10; mt i; long int ix /cli'scr();(1)；填写的语句：n=0 scanf(”s",s); for(i=0;i<strlen(s);i-H-)n=n* 10+(2);填写的语句:i+1pmitf(,n=%idn,3i)r7、从键盘输入一个正整数几计算并输

14、出n的阶乘。/include <math.h>#iiiclude <coiiio.h>#iiiclude <stdio.h> long fac( iiit n) mt i;long t=l;foi( i=2;i<=ii;i-H-) t = t*i;retimi (t);void main() int n; long iiit s; / ckscr();piiiitf(nEntei ail integer munber:M);9Yi平下平平平千平于平平平平Tcim/ |于平平千平于平平千于于于/(1);填写的语句:scaiif(n%ldr&n);

15、s = fac(n);f/千于千千千千平千千千干存TcimCi于千干千于于千千千于于千/%ldprintf(”The result is :(2)n",s);填写的语句: 三、编程题：(40分)1按下面的公式求S的近似值，直到最后一项值的绝对值小于X为止，X的 值由键盘输入，结果保留4位小数。分析：这是一个求若干个数之和的累加问题，终止循环的条件是最后一 项值的绝对值小于x, x的值由键盘输入，当退出循环时，sum的值己经 累加了最后一项a的值，因此要减去最后一项a的值。S=ll/3+l/5l/9+.4-(-l)A(n-l)l/2n-l/include "stdio.h&q

16、uot;#iiiclude "math.h”#mclude "stdlib.h"double fiui(double x)/水水水水水水水水水水*begjji水水水*水水水水水水水/ iiit y,i=l;double a、sum=0.0：do if(i%2=0) y=-i;else y=l; a=y*1.0/(2*i-l); sum=sum+a;iH;wliile (fabs(a)> =x); retiun siun-a;9YI/千于于于千千于于节于于Qllfl 千千千平于千于于于于于/ void main()double x,sum：printf(” E

17、nter x:nH);scanf(“lf； &x);sum=fim(x);print f(" sum=% .4 fn", sum); NONO();NONO()FILE *fi*fw;iiit i;double x;fi=fopenCD:exam03200310PROGIN.DAT”,T');fv=fopen(” D:exam03200310PROGOUT.DAT”,”w”); fdr(i=l;i<=5;i-H-) fscaiifi'/lf&x);fpniitf(fw,Ms=%.4firfim(x);fclose(fr);fclose(

18、fw); 2、将一个4位的整数m拆开，然后按相反的顺序输出，每输岀一位数其 后留2个空格。如m=1234,输出的结果应为m=1234,4 3 2 1程序如 下：（同学们自己编程，程序放在/*begiii*/ U/*end*/J2.间）#iiiclude "stdio.h"#iiiclude nmath.hM #iiiclude "stdlib.h" void main() iiit m=l234,a,b,c,d;piiiitf(nm=%d,%d %d %d %diVm,a,b,c,d); NONO(m,a,b,c,d);NONO( x,a,b,c,d)i

19、nt x,a,b,c,d：FILE *f;Ufopen(”D:exam03200302PROGOUT.DAT”,”w”); fpiiiitf(f,'x=%d: %3d%3d%3d%3d#n",x,a,b,c,d); fclose(f);3、按下面的公式,求S的近似值。S=l+l/2*2+l/3*3+.+l/n*n #iiiclude "stdio.h"#include Mmath.hH #iiiclude "stdlib.h" double fiui(iiit n)iiit i;double p,siun=0.0; for(i=l;i&

20、lt;=n;i-H-)sum=sum+l .0/p;retimi sum;void maiii() double s;iiit lx/cliscr();piiiitf( "Enter s:n");scaiif("%dM,&n);s=ftm(n);printf(”s=%.3fn”,s);NONO();NONO() FILEiiit i,m；fi=fopen(”D:exam03200306PROGIN.DAT”,T); fw=fopen(nD:exain03200306PROGOUT.DATM,wn); fbi(i=l;i<=5;i-H-) fscanf

21、(ft,"%d”,&m);fclose(fr);fclose(fw);4、求一个三数m的各位上的数字之和。m由键盘输入。（同学们#iiiclude <stdio.h> iiit fiiii(iiit m) void maiii() iiit in,s;/clrscr(); piiiitf(nEnter m :n”);scaiif(n%dr&m);s=fuii(m);piiiitf("Resiilt is: %dn",s);NONO();NONO() FILE *fr,*fw;iiit i,j,k,m：fi=fopen(”D:exam03

22、200308PROGIN.DAT”,”T”)； fw=fbpen(,D:exain03200308PROGOUT.DATn,wM);foi(i=l;i<=5;i-H-) fscaiif(fr,n%d",&m);垃 nntf(fv,"Sum = %dn”,fun(m);fclose(fr);fclose(fw);5、求两个正整数m和n的最大公约数，将求出的最大公数存入变 量t中。#iiiclude "stdio.h"#iiiclude "math.h"#include "stdlib.h"void main() int m=764i=40,t,i;/水水水水水*水水水*b egiii */9J/平