华为C语言机试题面试题汇总.doc

1第一题的题目大概是输入整型数组求数组的最小数和最大数之和，例如输入1,2,3,4则输出为5，当输入只有一个数的时候，则最小数和最大数都是该数，例如只输入1，则输出为2；另外数组的长度不超过50includestdio.hmain int num500; int i,n; printf请输入整型数组的长度150; scanfd,n; printf请输入整型数组的元素; for i0;in;i scanfd,numi; int min_numnum0; int max_numnum0; forint j0;jn;j ifmax_numnumj max_numnumj; else ifmin_numnumj min_numnumj; int summin_nummax_num; printf数组中最大与最小值之和dn,sum; return 0;2求两个长长整型的数据的和并输出，例如输入1233333333333333。3111111111111111111111111.。，则输出includestdio.hincludestring.hincludemalloc.hmain char *num1,*num2; 两个长长整型数据 char *sum; int temp;int len_num1,len_num2; 两个长长整型数据的长度 int len_max,len_min;num1char*mallocsizeofchar; num2char*mallocsizeofchar; printf输入两个长长整型数据; scanfs,num1; printf输入两个长长整型数据; scanfs,num2; len_num1strlennum1; len_num2strlennum2; len_maxlen_num1len_num2 len_num1len_num2; len_minlen_num1len_num2 len_num1len_num2;int len_max1len_max; sumchar*mallocsizeofchar*len_max;memsetsum,0 x00,len_max1;切忌初始化 for;len_num10len_num20;len_num1,len_num2 sumlen_maxnum1len_num1-1-0num2len_num2-1-0; iflen_num10 sumlen_maxnum1len_num1 - 1 -0; len_num1; iflen_num20 sumlen_maxnum1len_num2 - 1-0; len_num2; forint jlen_max1;j0;j 实现进位操作 tempsumj-0; ifsumj10 sumj-1sumj/10; sumj10; char *outsumchar*mallocsizeofchar*len_max1; j0; whilesumj0 跳出头部0元素 j;forint m0;mlen_max1;j,m outsummsumj0; outsumm0; printf输出两长长整型数据之和sn,outsum; return 0;3.通过键盘输入一串小写字母az组成的字符串。请编写一个字符串过滤程序，若字符串中出现多个相同的字符，将非首次出现的字符过滤掉。比如字符串“abacacde”过滤结果为“abcde”。要求实现函数void stringFilterconst char *pStr, long lLen, char *pOutputStr;【输入】pStr输入字符串lLen输入字符串长度【输出】pOutputStr输出字符串，空间已经开辟好，与输入字符串等长；include stdio.hincludestring.hincludemalloc.hvoid stringFilterconst char *p_str, long len, char *p_outstrint array2560;const char *tmp p_str;forint j0;jlen;j ifarraytmpj0 *p_outstr tmpj; arraytmpj; *p_outstr 0;void main char *str cccddecc char * outstr char *malloclen*sizeofchar; stringFilterstr,len,outstr; printfsn,outstr; freeoutstr; outstr NULL;4.通过键盘输入一串小写字母az组成的字符串。请编写一个字符串压缩程序，将字符串中连续出席的重复字母进行压缩，并输出压缩后的字符串。压缩规则1.仅压缩连续重复出现的字符。比如字符串abcbc由于无连续重复字符，压缩后的字符串还是abcbc.2.压缩字段的格式为字符重复的次数字符。例如字符串xxxyyyyyyz压缩后就成为3x6yz要求实现函数void stringZipconst char *pStr, long lLen, char *pOutputStr;【输入】pStr 输入字符串lLen 输入字符串长度【输出】pOutputStr 输出字符串，空间已经开辟好，与输入字符串等长；include stdio.hincludestring.hincludemalloc.hvoid stringZipconst char *p_str, long len, char *p_outstr int count1; forint i0;ilen;i ifp_strip_stri1 count; else ifcount1 *p_outstr count 0; *p_outstr p_stri; else *p_outstr p_stri; count 1;注意其位置 *p_outstr 0;void main char *str cccddecc压缩之前的字符串为sn,str; int len strlenstr; char * outstr char*malloclen*sizeofchar; stringZipstr,len,outstr; printf压缩之后的字符串为sn,outstr; freeoutstr; outstr NULL;5.通过键盘输入100以内正整数的加、减运算式，请编写一个程序输出运算结果字符串。输入字符串的格式为“操作数1运算符 操作数2”，“操作数”与“运算符”之间以一个空格隔开。补充说明1.操作数为正整数，不需要考虑计算结果溢出的情况。2.若输入算式格式错误，输出结果为“0”。要求实现函数void arithmeticconst char *pStr, long lLen, char *pOutputStr;【输入】pStr 输入字符串lLen 输入字符串长度【输出】pOutputStr 输出字符串，空间已经开辟好，与输入字符串等长；include stdio.hincludestring.hincludestdlib.hvoid arithmeticconst char *, long len, char *output char s110; char s210; char s310; int cnt 0; int len_strlen; forint i0;ilen_;i ifi cnt; ifcnt2 *output 0; *output 0; return; sscanf,s s s,s1,s2,s3; ifstrlens21s20s20- *output 0; *output 0; return; int len_s1strlens1; fori0;ilen_s1;i ifs1i0s1i9 *output 0; *output 0; return; int len_s3strlens3; fori0;ilen_s3;i ifs3i0s3i9 *output 0; *output 0; return; int x atois1; int y atois3; ifs20 int result xy;itoaresult,output,10; else ifs20- int result x-y; itoaresult,output,10; else *output 0; *output 0; return; void main char str 10 - 23 int len strlenstr; arithmeticstr,len,outstr; printfsn,str; printfsn,outstr;6.一组人(n个)，围成一圈，从某人开始数到第三个的人出列，再接着从下一个人开始数，最终输出最终出列的人(约瑟夫环是一个数学的应用问题已知n个人(以编号1，2，3.n分别表示)围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。)include stdio.hincludestring.hincludestdlib.hincludemalloc.htypedef struct Node int data; struct Node *next;LinkList;LinkList *createint n LinkList *p,*q,*head; int i1; pLinkList*mallocsizeofLinkList; p-datai; headp; fori1;in;i qLinkList*mallocsizeofLinkList; q-datai1; p-nextq; pq; p-nexthead; 使链表尾连接链表头，形成循环链表 return head; freep; pNULL; freeq; qNULL;void deletefunLinkList *L,int m LinkList *p,*q,*temp; int i; pL;whilep-nextp fori1;im;i qp; pp-next; printf5d,p-data; tempp; q-nextp-next; pp-next; freetemp; printf5dn,p-data;int main int n7,m3; LinkList *head1; head1createn; deletefunhead1,m; return 0;7输入一串字符，只包含“0-10”和“，”找出其中最小的数字和最大的数字(可能不止一个)，输出最后剩余数字个数。如输入“3,3,4,5,6,7,7”includestdio.hincludestdlib.hincludestring.hvoid main char str100; printf输入一组字符串n; scanfs,str; int lenstrlenstr; int array100; int count0; forint i0;ilen;i ifstri0stri9 arraycountstri-0; arraycount0; int resultcount; int minarray0; int maxarray0; forint j0;jcount;j ifmaxarrayj maxarrayj; else ifminarrayj minarrayj; forint k0;kcount;k ifarraykmin result; ifarraykmax result; printfdn,result;8.输入一组身高在170到190之间(5个身高)，比较身高差，选出身高差最小的两个身高；若身高差相同，选平均身高高的那两个身高；从小到大输出；如输入170 181 173 186 190输出170 173includestdio.hincludestdlib.hdefine N 5int main int HeightN; int dmin; int H1,H2; int i,j,temp; printf请输入一组身高在170到190之间的数据(共5个)n; forint k0;kN;k scanfd,Heightk; printfn; fori0;iN;i forj1;jN-iHeightj-1Heightj;j tempHeightj-1; Heightj-1Heightj; Heightjtemp; H1Height0; H2Height1; dminH2-H1; forint m2;mN;m ifHeightm-Heightm-1dmin H1Heightm-1; H2Heightm; dminHeightm-Heightm-1; printf身高差最小的两个身高为n; printfd,dn,H1,H2; return 0;9.删除子串，只要是原串中有相同的子串就删掉，不管有多少个，返回子串个数。include stdio.hinclude stdlib.hinclude assert.hinclude string.hint delete_sub_strconst char *str,const char *sub_str,char *result assertstr NULL sub_str NULL; const char *p,*q; char *t,*temp; p str; q sub_str; t result; int n,count 0; n strlenq; temp char *mallocn1; memsettemp,0 x00,n1; while*p memcpytemp,p,n; ifstrcmptemp,q 0 count; memsettemp,0 x00,n1; p p n; else *t *p; p; t; memsettemp,0 x00,n1; freetemp; return count;void main char s100 0; int num delete_sub_str“123abc12de234fg1hi34j123k”,”123”,s; printf“The number of sub_str is drn”,num; printf“The result string is srn”,s;10.要求编程实现上述高精度的十进制加法。要求实现函数void add const char *num1, const char *num2, char *result【输入】num1字符串形式操作数1，如果操作数为负，则num10为符号位-num2字符串形式操作数2，如果操作数为负，则num20为符号位-【输出】result保存加法计算结果字符串，如果结果为负，则result0为符号位。includestdio.hincludestdlib.hincludestring.h void movechar *str, int length 移除字母前的-符号 ifstr0 - return; int i; fori 0; i length-1; i stri stri1; stri 0;int remove_zerochar *result, int length int count 0; forint i length-1; i 0; i 从最后开始移除0，直到遇到非0数字，只对最初位置上的0不予判断 ifresulti 0 resulti 0; count; else return length-count; return length - count;void reversechar *result, int length 将字符串倒转 char temp; forint i 0; i length-1/2; i temp resulti; resulti resultlength-1-i; resultlength-1-i temp; int real_addchar *str1, char *str2, char *result, const bool flag int len1 strlenstr1; int len2 strlenstr2; int n1, n2, another 0; another表示进位 int cur_rs 0; 表示result的当前位数 int i, j; int curSum; fori len1-1, j len2-1; i 0 j 0; i, j n1 str1i - 0; n2 str2j - 0; curSum n1 n2 another; resultcur_rs curSum 10 0; another curSum / 10; ifj 0 whilei 0 遍历str1剩余各位 n1 str1i - 0; curSum n1 another; resultcur_rs curSum 10 0; another curSum / 10; ifanother 0 如果还有进位未加上 resultcur_rs another 0; else whilej 0 n2 str2j - 0; curSum n2 another; resultcur_rs curSum 10 0; another curSum / 10; ifanother 0 resultcur_rs another 0; resultcur_rs 0; cur_rs remove_zeroresult, cur_rs; ifflag resultcur_rs -;