福建省計算機二級考試C語言上機題庫改錯填空編程

1、福建省計算機二級測試C語言上機題庫改錯填空編程二、改錯題Cmody191.c修改程序,用以計算1100之間所有的偶數(shù)之和.#include<stdio.h>#include<math.h>voidmain()(inti=1;/*/intsum=2/*/;while(i+,i<=100)if(/*/i=(i*2/*/)continue;sum+=i;printf("Sumis%dn",sum);getch();Cmody192.c修改程序,將二維數(shù)組表示的方陣左下半三角(不含對角線)各元素加4,右上半三角(含對角線)各元素乘2.#include&

2、lt;stdio.h>#defineN5voidmain()intaNN,i,j;for(i=0;i<N;i+)for(j=0;j<N;j+)aij=i*5+j+11;printf("nArrayais:n");for(i=0;i<N;i+)for(j=0;j<N;j+)printf("%3d",aij);printf("n");)for(i=0;i<N;i+)/*/for(j=0;j<=i;j+)/*/aij+=4;for(i=0;i<N;i+)/*/for(j=i+1;j<N;

3、j+)/*/aij*=2;printf("nArrayaisturned:n");for(i=0;i<N;i+)for(j=0;j<N;j+)printf("%3d",aij);printf("n");)getch();Cmody201.c修改程序,計算滿足以下條件的整數(shù)n的最大值.22+42+62+82+2<1000#include<stdio.h>#include<math.h>voidmain()intn=0,sum=0;/*/while(sum<1000);/*/*/sum+=n

4、A2/*/;n+,n+;)printf("n=%dn",n-2);getch();)Cmody202.c修改程序,實現(xiàn)從鍵盤輸入一串字符,并在字符串中從第m個字符串開始截取n個字符.例如：輸入的字符串為“welcomed,m=2,n=3,那么截取的字串為“elc#include<stdio.h>#include<string.h>voidmain()charstr100,sub100;intm,n,len,i;printf("Enterstring:");gets(str);printf("nEntermn:"

5、);scanf("%d%d,&m,&n);len=strlen(str);if(m-1+n>len)printf("Can'trunwith%dand%d!n",m,n);elsei=0;/*/while(i<=n)/*/*/subi=strm+i;/*/i+;/*/subi='n'/*/printf("substringis:%sn",sub);getch();Cmody211.c修改程序,實現(xiàn)輸入一整型二維數(shù)組,計算其中數(shù)組元素的最大值與最小值的差.#include<stdio.h&

6、gt;#include<math.h>#defineROW3#defineCOL4voidmain()intaROWCOL,max,min,i,j,result;clrscr();printf("Enterarraya:n");for(i=0;i<ROW;i+)for(j=0;j<COL;j+)scanf("%d",&aij);/*/max=min=0;/*/for(i=0;i<ROW;i+)for(j=0;j<COL;j+)if(/*/min<aij/*/)min=aij;if(aij>max)/

7、*/max=aij/*/)printf("Result=%dn",max-min);getch();Cmody212.c修改程序cmody212.c,實現(xiàn)從鍵盤輸入一串字符和一個字串,輸出該子串在原字符串中出現(xiàn)的次數(shù).#include<stdio.h>#include<string.h>voidmain()(inti,j,k,count;chars1100,s2100;printf("EntermainString:");gets(s1);printf("EnterSubString:");gets(s2);c

8、ount=0;/*/for(i=0;*s1!='0'i+)/*/*/for(k=0;(s1j=s2k)&&(s1j!='0');j+,k+);/*/if(s2k='0')count+;printf("nCount=%dn",count);getch();Cmody221.c修改程序,輸出存儲在結(jié)構(gòu)體數(shù)組中8位學(xué)生成績的平均分以及最高學(xué)生的姓名#include<stdio.h>voidmain()(structstudentstu8="Mary",76,"John&quo

9、t;,85,"Tom,81,"Susa",87,"Wilu",79,"Yili",65,"Sonmu,73,"Lichar",70;inti=0,mrk;floattotal=0,aver,max;max=stu0.score;mrk=0;while(i<8)/*/total=total+stui;if(stui.score>max)max=stui.score;mrk=i;/*/i+/*/*/aver=total/8;printf("naver=%.2fnBestis%s

10、n,aver,);getch();Cmody222.c修改程序,完成其中n個字符串的升序排列/*/structurestudent/*/#include<stdio.h>charname10;floatscore;#include<string.h>#defineROW5#defineCOL30voidfun(charsCOL,intn)chartempCOL;inti,j;for(i=0;i<n-1;i+)/*/for(j=i+1;j<n-1;j+)/*/if(strcmp(si,sj)>0)(strcpy(temp,si);s

11、trcpy(si,sj);strcpy(sj,temp);)voidmain()(inti;charstrCOL="DEF","BAEELCS7FHIAME","CBADF","APMAE"/*/fun(str,COL);/*/clrscr();for(i=0;i<ROW;i+)printf("%sn",stri);getch();Cmody231.c修改程序cmody231.c,實現(xiàn)將輸入的十進(jìn)制正整數(shù)轉(zhuǎn)換成十六進(jìn)制數(shù),且用字符串存放該十六進(jìn)制數(shù)#include<math.h>

12、;#include<string.h>#include<stdio.h>voidmain()intx,b,i,j;chars5;printf("Inputanumber(Dec):");scanf("%d",&x);/*/i=1;/*/while(x>0)b=x%16;if(b>9)si=b-10+'A'else/*/si=b/*/;x=x/16;i+;printf("nHexnumberis:");for(j=i-1;j>=0;j-)putchar(/*/si/*/)

13、;getch();Cmody232.c修改程序cmody232.c,輸出以下9行數(shù)字金字塔112112321123432112345678987654321#include<stdio.h>voidmain()inti,j,k;for(k=1;k<=9;k+)/*/for(i=0;i<=k;i+)/*/printf("");for(i=1;i<=k;i+)printf(''%cT,i);/*/for(j=k-1;j>1;j-)/*/printf("%cT,j);printf("n");)get

14、ch();)a+b>c/*/*/*/bigmedian三、填空題Ccon191.c程序ccon191.c,從鍵盤輸入3個實型值,假設(shè)它們能構(gòu)成一個三角形即任意兩邊之和大于第三邊,那么根據(jù)以下公式求該三角形面積；否那么,提示相應(yīng)信息.aread(da)(db)(dc)abc其中：d2#include<stdio.h>#include<math.h>voidmain()floata,b,c,d,area;printf("Pleaseinputnumbers:n");scanf("%f,%f,%f",&a,&b,&

15、amp;c);if(/*/)d=(a+b+c)/2;area=/*/;printf("areais%.2fn",area);)elseprintf("NotTriangle!n");getch();Ccon192.c程序ccon192.c,計算：大貨車限載8噸、中型貨車限載3噸、微型貨車限載1噸,需50輛車運輸100噸貨物時,應(yīng)配備的大、中、微型車分別多少輛？共有多少種解決方案？#include<stdio.h>#include<conio.h>int/*/()intbig,median,small,n=0;printf(&quo

16、t;smalln");for(big=0;big<=13;/*/*/)for(median=0;median<=33;median+)small=100-8*big-3*median;if(/*/*/=50)n+;printf(''%d->%2d%2d%2dn",n,big,median,small);)returnn;)voidmain()intnum;num=fun();printf("nThereare%dsolutions.n",num);getch();)*p2=sum-*p1;Ccon201.c程序ccon2

17、01.c)求一堆蘋果的總數(shù),之間.假設(shè)按每袋裝根據(jù)以下袋裝情況總數(shù)在1003008個蘋果那么多出5個,假設(shè)按每袋裝7個蘋果,那么多出4個,假設(shè)按每袋裝5個蘋果,那么多出2個.#include<stdio.h>voidmain()intn;for(n=300;/*/voidmain()intx,y;printf("pleasenumbers:n");scanf("%d,%d",&x,&y);printf("nOriginal:y=%dn",x,y);swap(&x,&y);printf(&qu

18、ot;nNow:y=%dn",x,y);getch();)inputx=%dx=%d/*/;n-)if(n%8=5)if(!(n-4)%7)/*/Ccon211.c程序ccon211.c,按每行最多5個數(shù)/*/)printf("n=%dn",n);)getch();)方式輸出100400之間能同時被7和4整除的所有數(shù)及它們之和.#include<stdio.h>voidmain()Ccon202.c程序ccon202.c中的swap函數(shù),通過指針變量實現(xiàn)兩個變量值的交換.inti,n;intsum=0;/*/*/#include<stdio.h&

19、gt;void/*/)intsum;sum=*p1+*p2;*p1=/*/for(i=101;i<=399;i+)/*/;swap(/*/if(/*/*/&&!(i%4)printf("%dsum+=/*/,i);/*/;n+;if(n%5=0)printf("n");)printf("nsum=%dn",sum);getch();)Ccon212.c程序ccon212.c中的cmob函數(shù)用以輸出等式a*b*c+a*c*c=560的所有a,b,c的組合(其中a,b,c只取19之間的一位整數(shù)).如a=7,b=2,c=8時,7

20、*2*8+7*8*8=560#include<stdio.h>voidcomb(/*/*/)inta,b,c;for(a=1;a<10;a+)for(b=1;b<10;b+)for(c=1;c<10;c+)(if(/*/*/=m)printf("a=%d,b=%d,c=%dn",a,b,c);)voidmain()(intresult=560;comb(result);getch();Ccon221.c程序ccon221.c,輸出如以下圖形%&&&&&&%&&&&%

21、&&%#include<stdio.h>#include<conio.h>voidmain()(introw,col;for(row=1;row<=7;row+)printf("t");for(col=1;col<=/*/*/;col+)if(row%2)printf("%c'',/*/*/);elseprintf("%c'',/*/*/);printf("n");)getch();)Ccon222.c程序ccon222.c中的pnt函數(shù),根據(jù)參數(shù)n(

22、2&n&10)輸出類似如下所示的方針(n=5).123451098761112131415pnt(/*/*/,n);getch();)20191817162122232425#include<stdio.h>#include<conio.h>#defineM10Ccon231.c程序ccon231.c,計算以下分?jǐn)?shù)序列的前18項之和235813,12358voidpnt(intaMM,intn)inti,/*/*/;for(i=0;i<n;i+)for(j=0;j<n;j+)if(i%2=0)a皿尸/*/*/;elseaij=i*n+n-j;

23、)printf("arraya(n=%d)is:n",n);for(i=0;i<n;i+)for(j=0;j<n;j+)printf("%3d",aij);printf("n");)voidmain()intaMM;intn;printf("Inputn(2<=nAndn<=10):");scanf("%d",&n);#include<stdio.h>voidmain()fun(x)50.010.008(x2000)44x500200044x500x2

24、000x50005000inti;floatsum,a=2,b=1,t;/*/*/;for(i=1;i<=18;i+)sum+=/*/t=a;a=a+b;b=/*/*/;)printf("sum=%fn",sum);getch();)/*/;Ccon232.c程序ccon232.c中的inv函數(shù),通過指針變量將字符串str的內(nèi)容asdfjk調(diào)用inv函數(shù)后,變成："kjfdsa."#include<stdio.h>#include<string.h>/*/voidinv(char*str)inti,j,k;j=/*/*/;f

25、or(i=0;i<strlen(str)/2;i+,j-)k=*(str+i);*(str+i)=*(str+j);*(str+j)=/*/*/;/*/voidmain()charstr="asdfjk"printf("nOriginalis:%sn",str);inv(str);printf("nNowis:%sn",str);getch();四、編程題cprog191.c翻開程序cprog191.c,完成其中的fun函數(shù)：根據(jù)輸入的匯款數(shù)額,求應(yīng)交的匯費.設(shè)應(yīng)交匯費的計算公式如下：#include<stdio.h>

26、;#include<math.h>doublefun(floatx)voidmain()floatx;doubley;clrscr();printf("Pleaseinputanumber:n");scanf("%f",&x);y=fun(x);printf("f(%.2f)=%.2fn,x,y);getch();cprog192.c翻開程序cprog192.c,完成其中的fun函數(shù)：對4義4矩陣從頂行開始各行按從左到右順序查找,找出各行中0元素之后的所有負(fù)數(shù),并按元素出現(xiàn)的先后順序存放到數(shù)組b中,并返回這些負(fù)數(shù)之和.如矩陣

27、為：2031820403312210133那么調(diào)用函數(shù)fun()后,數(shù)組b各元素依次為-3,-1,-4,-3,-12,-13,并返回-36.#include<stdio.h>#include<math.h>#defineROW4#defineCOL4#defineLEN12intfun(intaCOL,intb)/*/getch();)cprog201.c翻開fun)cprog201.c,完成其中的函數(shù)該函數(shù)的數(shù)學(xué)表達(dá)式是1/(1ex)x10fun(x)(x3)sin(2x)x10#include<stdio.h>#include<math.h>

28、doublefun(floatx)/*/*/inputvoidmain()intarraROWCOL=-2,0,-3,-1,-8,2,0,-4,0,3,-3,-12,21,0,-13,3;intarrbLEN,i,result;for(i=0;i<LEN;i+)arrbi=0;clrscr();result=fun(arra,arrb);printf("arrayarrbis:n");for(i=0;i<LEN;i+)if(arrbi=0)printf("");elseprintf("%4d",arrbi);printf(

29、"nresult=%d",result);/*/voidmain()floatx;doubley;printf("Pleasenumber:n");scanf("%f",&x);y=fun(x);printf("f(%.3f)=%.3fn'',x,y);getch();)cprog202.c翻開程序cprog202.c,完成其中的fun函數(shù)：函數(shù)fun(doublex,intn)式近似計算cos(x)的值:2462nxxxnxcos(x)1(1)2!4!6!(2n)!#include<stdio

30、.h>voidmain()doublex;intn;doublefun(doublex,intn);printf("Pleaseenterx,n:");scanf("%lf%d",&x,&n);#include<stdio.h>#include<math.h>doublefun(floatx)/*/printf("cos(%lf)=%lfn",x,fun(x,n);getch();)doublefun(doublex,intn)(/*/*/)voidmain()floatx;doubley

31、;clrscr();printf("Pleasenumber:n");scanf("%f",&x);y=fun(x);input/*/)cprog211.c翻開fun)fun(x)cprog211.c,完成其中的函數(shù)該函數(shù)的數(shù)學(xué)表達(dá)式是ln(3x)x102x11x10x1printf("f(%.2f)=%.2fn",x,y);getch();)cprog212.c翻開程序cprog212.c,完成其中的fun函數(shù)：對兩個字符串中小寫英文字母的個數(shù)進(jìn)行比擬,并返回比擬的結(jié)果.(注：第1個字符串比第2個字符串的小寫英文字母個數(shù)多,

32、那么返回正數(shù),相等返回0,個數(shù)少那么返回負(fù)數(shù))ofofLowerLowerofofLowerLower例如：字符串str1為“abc123字符串str2為abcd'；那么函數(shù)fun的返回值應(yīng)小于0.#include<stdio.h>intfun(char*str1,char*str2)/*/printf("Numberscharin%s=Numberscharin%s,str1,str2);if(rela<0)printf("Numberscharin%s<Numberscharin%s,str1,str2);getch();/*/voidm

33、ain()intrela;char*str1,*str2;clrscr();printf("PleaseNO.1:");gets(str1);printf("PleaseNO.2:");gets(str2);rela=fun(str1,str2);if(rela>0)printf("Numbersinputinputofcharin%s>Numbersofcharin%s,str1,str2);if(rela=0)stringstringLowerLowercprog221.c翻開程序cporg221.c,完成其中的fun(x,y)

34、(y2)xy1.0x2.0,0y6fun函數(shù)：根據(jù)輸入的x和y的值,按如下公式計算：將計算結(jié)果的整數(shù)局部存入數(shù)組元素arr0,并將該計算結(jié)果的小數(shù)部分的前兩位存入arr1中.#include<stdio.h>#include<math.h>voidfun(intarr,doublex,inty)/*/*/voidmain()inty;f(x,n)doublex;intarra2;printf("Pleaseenterx(1<x<2),y(0<y<=6):n");scanf("%lf%d",&x,&a

35、mp;y);if(x>1.0&&x<2.0)&&(y>0&&y<=6)fun(arra,x,y);printf("npart1=%d,part2=%d",arra0,arra1);elseprintf("xoryisoutofrange!");getch();)cprog222.c翻開程序cprog222.c,完成其中的fun函數(shù)：將一個升序數(shù)組a和一個降序數(shù)組b中的所有元素按降序存入數(shù)組c中.例如：數(shù)組a為5,10,15,27,46,數(shù)組b為50,45,42,29,15,8,5,2

36、,那么數(shù)組c就應(yīng)為50,46,45,42,29,27,15,15,10,8,5,5,2.#include<stdio.h>#defineN5#defineM8voidfun(inta,intb,intc)/*/*/voidmain()intarraN=5,10,15,27,46,arrbM=50,45,42,29,15,8,5,2;intarrcN+M,k;fun(arra,arrb,arrc);printf("arrayarrais:");for(k=0;k<N;k+)printf("%d",arrak);printf("n

37、arrayarrbis:");for(k=0;k<M;k+)printf("%d",arrbk);printf("nresultarrcis:");for(k=0;k<N+M;k+)printf("%d",arrck);getch();cprog231.c翻開程序cprog231.c,完成其中的函數(shù)fun該函數(shù)的遞歸公式是:xn1(x1)f(x,n1)n1#include<stdio.h>#include<math.h>voidmain()(doublex;intn;doublefun(d

38、oublex,intn);printf("Pleaseenterx,n:n");scanf("%lf%d",&x,&n);printf("fun=%lfn",fun(x,n);getch();doublefun(doublex,intn)(/*/*/)cprog232.c翻開程序cprog232.c,完成其中的fun函數(shù)：返回一維數(shù)組中的次大數(shù)(即僅小于最大數(shù)的數(shù)),設(shè)數(shù)組a中沒有重復(fù)的元素.例如：數(shù)組元素為45,58,33,24,40,20,30,28,31,函數(shù)fun的返回值為45.#include<stdi

39、o.h>#defineROW9intfun(inta)/*/voidmain()intarraROW=45,58,33,24,40,20,30,28,31;intmax_pre,i;max_pre=fun(arra);printf("arrayarrais:");for(i=0;i<ROW;i+)printf("%d",arrai);printf("nmax_preis%d",max_pre);getch();/*/subi=strm+i;1*1susubi=strm-1+i;n'/*/fsubi曲cmody211

40、.c/*/max=min=0;/*/fmax=min=a00;/*/min<aij/*/fmin>aij/*/max=aij/*/fmax=aij;cmody212.c/*/for(i=0;*s1!='0'i+)/*/ffor(i=0;*(s1+i)!=0''i+)/*/for(k=0;參考答案(s1j=s2k)&&(s1j!='0');二、改錯題j+,k+);/*/-cmody191.cfor(k=0,j=i;/*/intsum=2/*/;-intsum=0;(s1j=s2k&&(s2k!=0'

41、;');j+,cmody192.c/*/for(j=0;j<=i;j+)/*/j<i;j+)/*/for(j=i+1/*/-for(j=i;j<N;j+)/*/intsum=2/*/;fif(i!=(i/2)*2)k+);cmody221.cffor(j/*astructurestudent/*/-structstudent;j<N;j+)/*/total=total+stui;/*/-total=total+stui.score/*/i+/*/fi+;cmody201.c/*/while(sum<1000);/*/fwhile(sum<1000)/

42、*/sum+=nA2/*/;/*/-sum+=n*ncmody222.c/*/for(j=i+1;j<n-1;j+)/*/-for(j=i+1;j<n;j+)/*/fun(str,COL);/*/ffun(str,ROW);while(i<n)Cmody231.ccmody202.c/*/while(i<=n)/*/*/i=1;/*/fi=0;/*/si=b/*/Cmody232.csi=b+'0'if(/*/!(i%7)&&!(i%4)sum+=/*/i/*/;/*/*/*/*/for(i=0;i<=k;i+)>for(i=

43、0;i<9-k;i+)for(j=k-1;j>1;j-)/*/ffor(j=k-1;j>=1;j-)三、填空題Ccon191.cif(Ccon212.cvoidcomb(/*/intmif(/*/a*b*c+a*c*c=m)/*/)/*/a+b>c/*/&&b+c>a&&c+a>b/*/)area=/*/sqrt(d*(d-a)*(d-b)*(d-c)/*/Ccon192.cint/*/fun/*/()for(big=0;big<=13;/*/big+/*/)Ccon221.cfor(col=1;col<=/*/*

44、/;col+)printf("%c",/*/*/);printf("%c",/*/*/);rowif(/*/*/=50)big+median+smallCcon222.cinti,/*/j/*/;a皿尸/*/i*n+j+1pnt(/*/a/*/,n);/*/;Ccon201.cfor(n=300;/*/;n-)if(!(n-4)%7)=0/*/)/*/n>=100/*/&&(n-2)%5=Ccon231.c/*/sum=0sum+=/*/a/bb=/*/t/*/;/*/;/*/;Ccon202.cvoidswap(/*/int*p1

45、,int*p2/*/)*p1=/*/*p2/*/;Ccon232.cj=/*/strlen(str)-1/*/;*(str+j)=/*/k/*/;Ccon211.c/*/n=0/*/四、編程題cprog191.cx-1)；doublefun(floatx)/*/doubley;if(x<=500)y=5;elseif(x<=2000)y=0.01;elseif(x<=5000)y=0.008*(x-2000);elsey=44;returny;/*/cprog192.cintfun(intaCOL,intb)/*/inti,j,k,t=0,sum=0;for(i=0;i<ROW;i+)for(j=0;j<COL;j+)if(aij=0)for(k=j+1;k<COL;k+)if(aik<0)bt=aik;sum+=bt;t+;returnsum;/*/el