編程題目-c語言程序

信盈達(dá)就業(yè)班學(xué)員面試過程中常見C筆試題C語言編程題10100268后再開方，如果開方后的#include#include<stdio.h>#includeint{longintfor(i=1;i<100000;{x=sqrt(i+100);/*x100后開方后的結(jié)果*/y=sqrt(i+268)*y168后開方后的結(jié)果*/)/方數(shù){}}}355#include<stdio.h>#include<math.h>int{intday,month,year,sum,leap;{casecasecasecasecasecasecasecasecasecasecasecase}/if(year%400==0||(year%4==0&&year%100!=0)*判斷是不是閏年*/if(leap==1&&month>2*2,總天數(shù)應(yīng)該加一天*/printf("Itisthe%dthday.",sum);}程序分析：想辦法把最小的數(shù)放到x上，先將x與y進(jìn)行比較，如果x>y則將x與y的值進(jìn)行交換，xzx>zxzx最小。#include#include<stdio.h>#includeint{intx,y,z,t;if(x>y){ /*x,y的值*/}{ }{ }printf("smalltobig:%d%d%d\n",x,y,z);}9*999列，i控制行，jintmain(void){inti,inti,for(i=1;i<=9;{//for(j=1;j<9;j++)//口訣形式for(j=1;j<=i;j++) {printf("%d*%d=%-3d",j,}}return0;}intmain(void){longf1,f2;inti;for(i=1;i<=20;{{{ };/}return0;}intmain(void){intm,i,k,h=0,leap=1;for(m=101;m<=200;{for(i=2;i<=k;{}{{}}}printf("\nThetotalis%d",h);return}如：153153=1的三次方＋5的三次方＋3的三次方。intmain(void){intprintf("'waterflower'numberis:");for(n=100;n<1000;n++){{}}return0;}nk，然后按下述步驟完成：(1)n，則說明分解質(zhì)因數(shù)的過程已經(jīng)結(jié)束，打印出即可。如果n<>k，但n能被k整除，則應(yīng)打印出k的值，并用n除以k的商,作為新的正整數(shù)你n, nkk+1k的值,intmain(void){intprintf("\npleaseinputanumber:\n");for(i=2;i<=n;{{}{}}return0;}題目：利用條件運算符的嵌套來完成此題：學(xué)習(xí)成績>=90A表示，60-89B表示，60分以下的用C表示。intmain(void){intscore;printf("pleaseinputascore\n");printf("%dbelongsto%c",score,grade);return0;}intmain(void){inta,b,num1,num2,temp;printf("pleaseinputtwonumbers:\n");{}while(b!=0)/*b0為止{}return}intmain(void){charintletters=0,space=0,digit=0,others=0;printf("pleaseinputsomecharacters\n");{{}elseif(c=='{}else{}{}}printf("allinall:char=%dspace=%ddigit=%dothers=%d\n",letters,space,digit,others);return0;}題目求s=a+aa+aaa+aaaa+aa...a的值其中a是一個數(shù)字例如2+22+222 (此時共有5個數(shù)相加)，intmain(void){inta,n,count=1;longintsn=0,tn=0;printf("pleaseinputaandn\n");{}returnreturn}intmain(void){staticintk[10];inti,j,n,s;for(j=2;j<1000;{for(i=1;i<j;{{}}{printf("%d是一個完數(shù):",j);for(i=0;i<n;i++){}}}return0;}題目：一球從100米高度落下，每次后反跳回原高度的一半；再落下，求它在第10次時，共經(jīng)10次反彈多高？intmain(void){floatsn=100.0,hn=sn/2;intn;for(n=2;for(n=2;n<=10;{}printf("總共經(jīng)過%f米\n",sn);printf("10%f米\n",hn);return}10天早上想再吃時，見只剩下一個桃子intmain(void){{x1=(x2+1)*2;/*212倍}printf("thetotalisreturn0;}題目：兩個乒乓球隊進(jìn)行比賽，各出三人。甲隊為a,b,c三人，乙隊為x,y,z三人。已抽簽決定比賽。有人向隊員打聽比賽的。a說他不和x比，c說他不和x,z比，請編程序找出三隊賽手的。intmain(void){charintmain(void){chari,j,k;/*ia的對手，jb的對手，k是c的對手*/for(i='x';i<='z';i++)for(j='x';j<='z';{for(k='x';k<='z';{{{{}}}}return0;}orderisa-- c—intmain(void){chari,j,k;/*ia的對手，jb的對手，k是c的對手*/for(i='X';i<='Z';i++){for(j='X';j<='Z';{for(k='X';k<='Z';{{{}printf("RivalsofA,BandCare:%c%c}}}return0;}RivalsofA,BandCare:ZX**程序分析：先把圖形分成兩部分來看待，前四行一個規(guī)律，后三行一個規(guī)律，利重for循環(huán)，第一層int{intfor(i=0;i<=3;{for(j=0;j<=2-i;{printf("}for(k=0;k<=2*i;{}}for(i=0;i<=2;{for(j=0;j<=i;{printf("}for(k=0;k<=4-2*i;{}}return0;}intmain(void){ floata=2,b=1,s=0;{}printf("sumis%9.6f\n",s);return}intmain(void){floatn,s=0,t=1;for(n=1;n<=20;n++){}return0;}intfact(j);int{intintfact();for(i=0;i<5;i++){}}int{intsum;{}{}return}voidpalin(n);int{intvoidpalin(intprintf("5個字符，回車結(jié)束:");}void{charnext;{}{{}}依次類推，推到第一人（10歲，再往回推。intage(n);int{}int{intc;{}{}}intage(n);int{long 位if{printf("thereare5,%ld%ld%ld%ld}}elseif{printf("thereare4,%ld%ld%ld}elseif{printf("thereare3,%ld%ld}elseif{printf("thereare2,%ld}elseif{printf("thereare}}charis_palindrome(longlong intmain(void){longlong intret;printf("Pleaseinputainttypedigit:\n");scanf("%lld",&v);ret=is_palindrome(v);if(ret){printf("%lld是回文數(shù)字}{printf("%lld不是回文數(shù)字}} 無Called Return 真：是回文數(shù)字; dateof charis_palindrome(longlongnum){ints=longlong bNum=num;intmod; charwhile(bNum!=0){mod=bNum%10; //123%10=3s=s*10+mod; //s=0*10+3bNumbNum //bNum123/10=12(int自動轉(zhuǎn)換}return(s==}100#defineN intmain(void){inti,j,line,a[N];for(i=2;i<N;i++){a[i]=}for(i=2;i<sqrt(N);i++)for(j=i+1;j<N;j++){if(a[i]!=0&&a[j]!=if(a[j]%a[i]=={a[j]=}}for(i=2,line=0;i<N;{if(a[i]!={printf("%5d",a[i]);}if(line=={line=0;}}}10#defineN10int{int/*inputprintf("pleaseinputtennum:\n");for(i=0;i<N;i++){}for(i=0;i<N;i++){}/*sorttennum*/for(i=0;i<N-1;{for(j=i+1;j<N;j++){}}/*outputdata*/printf("Aftersorted\n");for(i=0;i<N;i++){}}題目：求一個3*3矩陣對角線元/*TC*/staticvoiddummyfloat(float*x){floaty;}int{floata[3][3],sum=0;inti,j;printf("9個整數(shù):\n");for(i=0;i<3;i++)for(j=0;j<3;{}for(i=0;i<3;{}printf("} int{inta[11]=printf("originalarrayis:\n");for(i=0;i<10;i++){}printf("insertanewnumber:");{}{for(i=0;i<10;{{for(j=i+1;j<11;j++){}}}}for(i=0;i<11;{}}#defineN5int{inta[N]={9,6,5,4,1},i,temp;printf("\noriginalarray:\n");for(i=0;i<N;i++){}for(i=0;i<N/2;{}printf("\nsortedarray:\n");for(i=0;i<N;i++){}}題目：學(xué)習(xí)staticvoid{intstaticintstatic_var=0;printf("\40:varequal%d\n",var);printf("\40:staticvarequal%d\n",static_var);} {intfor(i=0;i<3;{}}}auto {inti,num;for(i=0;i<3;{printf("\40:Thenumequal%d\n",num);{autointprintf("\40:Theinternalblocknumequal%d\n",num);}}}1 {intintfor(i=0;i<10;i++){}for(i=2;i<10;i++)for(j=1;j<i;j++){}for(i=0;i<10;{for(j=0;j<=i;{}}}voidswap(p1,p2); {intint*pointer1,*pointer2,*pointer3;printf("pleaseinput3number:n1,n2,n3:");{}{}{}printf("thesortednumbers}voidswap(int*p1,int{intp;}}#defineN7voidtest(intarr[],intint{intarr[N1356497};//}voidtest(intarr[],int{intmax0min0;//inti10i20;//inttemp0;//inti=for(i=0;i<num;{printf("%d",arr[i]}maxminarr[0];//for(i=1;i<num;{ifmax //{max=arr[i];i1=i;}if(min>{min=arr[i];i2=i;}}//temp=arr[0];arr[0]=arr[i1];arr[i1]=*if(min=={i2=}//temp=arr[num-1];arr[num1arr[i2];arr[i2]=temp;printf("%d===printf("%d===for(i=0;i<num;{printf("%d",arr[i]}}nmmm#defineN7voidmove(intarr[],intn,intint{intnumber[20],n,m,i;printf("請輸入%d個整數(shù):",n);for(i=0;i<n;i++){}for(i=0;i<n;i++){}}voidmove(intarray[],intn,int{int*p,array_end;{}{}}n個人圍成一圈，順序排號。從第一個人開始報數(shù)（13報數(shù)3的人退出#defineNMAX50int{intprintf("pleaseinputthetotalofnumbers:");for(i=0;i<n;{{{}{{}}{}printf("%dprintf("%disleft\n",*p);}題目：實現(xiàn)標(biāo)準(zhǔn)Cintmystrcmp(const char*str1,constchar*str2);intmain(void){intii=mystrcmp("123a","123ab");printf("%d\n",i);i=mystrcmp("123ab","123a");printf("%d\n",i);i=mystrcmp("123b",i=mystrcmp("123a",i=mystrcmp("123a",printf("%d\n",i);}int char*str1,constchar{//inti=//while(str1[i]||str2[i])while(*str1||*str2){//if(str1[i]!=str2[i])if(*str1!=*str2){} }//return(str1[i]-//return(*--str1 *--str2);//如果上面if(*str1++!=*str2++)return(*str1-*str2); //如果上面if(*str1!=*str2)}#defineCOUNTstructNODE{unsignedlongchar structNODE 無 Return dateof structNODE*createNode(intn){structNODE*pHeadNULL*pRear // = inti;intret;char*strName=(charprintf("\r\n16字節(jié)以內(nèi)！\r\n");for(i=0;i<n;i++){pNew=(structNODE*)malloc(sizeof(struct{printf("IDret=scanf("%d%15s",&pNew->uID,if(ret!={fflush(stdin);//清緩沖 } if(getchar()!={printf("15字節(jié)，請重新輸入！\r\n");}}whileif0{pHead=}{pRear->next=}pNew->next=pRear=}return} 無Called pHead鏈表頭 Return dateof voidprintNode(structNODE*pHead){inti=structNODE*pTemp=if(pHead==NULL){}while(pTemp!={printf("節(jié)點%-03dID:%-03dname:%-10snipTemp->uIDpTemp->strName);pTemp=pTemp->next;}}int{structNODEpHead=createNode(COUNT);}/*reverseoutputa#include"stdlib.h"#include"stdio.h"#include"conio.h"structlist{intstructlisttypedefstructlistnode;typedefnode*link; {intnum,i;printf("\npleaseinput5data==>\n");for(i=0;i<=4;i++){}{{printf("Thevalueis==>%d\n",ptr->data);}}struct{intstructlisttypedefstructlistnode;typedefnode*link;voidprint_node(link{linktmp;{printf("%d",tmp->data);}}{linktmp1,tmp2,pointer;inti;for(i=1;i<num;{}return}link inkpointer1,link{linktmp;{}returnpointer1;}int{intarr1[]=intarr2[]=linkptr1,ptr2;ptr1=concatenate(ptr1,ptr2);}#defineN4voidswap(char*p1,char*p2);intmain(void){charprintf("pleaseinputthreestrings\n");{}{}{}printf("afterbeingsorted\n");}voidswap(char*p1,char{char*p[20];} {inti,m,j,k,for(i=4;i<10000;i+={count=m=for(k=0;k<5;{j=i/4*5+1;i=j;if(j%4=={}{}}i=if(count=={printf("%d\n",count);}}}題目：809*??=800*??+9*??+1其中??代表的兩位數(shù),8*??的結(jié)果為兩位數(shù)，9*??3位數(shù)。求??代表809*??后的結(jié)果。voidoutput(longb,long{printf("\n%ld/%ld=809*%ld+%ld",b,i,i,}int{longinta,b,i;a=809;for(i=10;i<100;{b=i*a+if(b>=1000&&b<=10000&&8*i<100&&9*i>=100)output(b,i);}}int{char*p,s[6];intn;p={}}0,1,2,3,4,5,6,714個。27*4個。37*8*4個。47*8*8*4個。int{longsum=4,s=4;intj;for(j=2;j<=8;j++)/*jisplaceof{{}{}}}voidtest(void);intmain(void){test()}void{intchararr[]={'a','b','c','d','e','f','g','h','i','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};for(i=0;i<strlen(arr);{for(j=strlen(arr);j>i;j--{printf("}forj=i;j>0j //{}printf("%c",arr[0]);//forj0ji; //{printf("%c",arr[j+}}printf("\n");//}}intmain(void){intfor(i=0;i<=3;i++){}for(i=0;i<=3/2;{}for(i=3;i>=0;i--{}return0;}intmain(void){intsum=0;printf("pleaseinputtwostrings\n");{{{}}{}{}}return} 由編譯器自動分配，存放函數(shù)的參數(shù)值，局部變量的值等 若程序員不，程序結(jié)束時可能由OS回收 {charchar*str=&a; return}沒有為str分配內(nèi)存空間，將會發(fā)生異常。問題出在將一個字符串進(jìn)一個字符變量指針?biāo)傅刂?。雖然可棧只能從頭部取數(shù)據(jù)也就最先放入的需要遍歷整個棧最后才能取出來，而且在遍歷數(shù)據(jù)的時候還得為數(shù)據(jù)#ifndefINCvxWorksh#defineINCvxWorksh#ifdefcplusplusextern"C"{#ifdef}#endif/*INCvxWorksh INCvxWorksh INCvxWorksh名字與C語言的不同。例如，假設(shè)某個函數(shù)的原型為：voidfoo(intx,inty);該函數(shù)被Csymbol庫中為了實現(xiàn)C和C++的混合編程，C++提供了C連接交換指定符號extern"C"來解決名字匹配問題，函數(shù)前加extern"C"后，則編譯器就會按照C語言的方式將該函數(shù)編譯為_fooC語言中就可以調(diào)用C++的函數(shù)了。結(jié)構(gòu)和聯(lián)合都是由多個不同的數(shù)據(jù)類型成員組成,但在任何同一時刻,對于聯(lián)合的不同成員賦值,將會對其它成