C + + Some pen questions

#include <iostream> #include <math.h> #include <string.h>using namespace std;// Cow birth: A baby cow can give birth to the first cow in the 4th year   after a year//existing a baby cow  20 years   Total number of cows int cownum (int size) {int*  Year = new int[size];memset (year, 0, size*sizeof (int)); year[0] = 1;int  count = 0;for  (int i = 0; i < size; ++i) {if  (i  >= 4-1 ) {year[i] = year[i - 1] + year[i - 3];}} for  (int i = 0; i < size; ++i) Count += year[i];return  count;delete year;} INT&NBSP;COWNUM1 (int year) {int newcow1 = 1;//1   ox    start with 1 int newcow2 &NBSP;=&NBSP;0;//2   Ox int newcow3 = 0;//three years   Ox int newcow4 = 0;//four years   cattle    Adult Cow//int newcow5 = 0;//will test instructions understand as the 5th year the cow begins to live    add a 5-year-old cattle number int oldcow =  0;//Total for  (int yearcnt = 2; yearcnt ), Cub   Cow int count = 0; // &LT;=&NBSP;YEAR;&NBSP;++YEARCNT)//Default ox is 1 years old   first time to enter the ox  2 years old {//newcow5 = newcow4;newcow4 = &NBSP;NEWCOW3;&NBSP;//3 Longhorn  4 year old oldcow += newcow4;newcow3 = newcow2;newcow2 =  newcow1;//one-year-old ox   2-year-old NEWCOW1&NBSP;=&NBSP;OLDCOW;//1-year-old cow   for the number of calves to be born count = oldcow + Total number of  newcow1+newcow2+newcow3;//is   adult ox  + 1 2 3 year old Ox//count = oldcow +  newcow1 + newcow2 + newcow3+newcow4; //if it starts at 5 years old//total   adult bovine  + 1  2 3 4-year-old bull}return count;} A string of beads   a m<=10 color  n   for the smallest substring with all colors//logic complex problem Void allcolorbead (char* bead, int  colornum)//bead  Bead string  colorNum  number of colors {char* head,* str=bead; //head  shortest sub-string head   str  cyclic variable Int clrnum = 0 ,beadcnt=0,tmpcnt=1000000,tmpclrnum=-1The char colorarr[128] = { 0 };//is used to determine whether each color first appears   to distinguish the number of colors +1while  (*str!= ') { if  (* (str + 1)  != *str)  //If the connected color jumps directly over {char* tmp = str; // Record str  If the substring meets the requirements  head=strwhile  ((*tmp)! = ' \ ") {if  (colorarr[*tmp] == 0)//color first appears &NBSP;CLRNUM++{++CLRNUM;++COLORARR[*TMP];} if  (clrnum == colornum)  //tell bead Color number logic//{//if  (TMPCNT&NBSP;&GT;&NBSP;BEADCNT)// If the substring meets the requirements  head = str  jumps out of the loop//{//tmpcnt = beadcnt;//head = str;//}//break ;//}if  (Tmpclrnum==clrnum)//Don't tell beads color number   First time need to fully traverse   to find out the number of colors break;++beadcnt;++tmp;} if  (Tmpcnt > beadcnt&&clrnum>=tmpclrnum)//If the substring satisfies the requirements  head = str   Jump out of the loop {tmpcnt = beadcnt;tmpclrnum = clrnum;//does not tell the bead color number needs to add the condition head = str;} clrnum = 0; //each retrieval substring End    counter system 0 Beadcnt = 0;memset (colorarr, 0, 128  *&nbsP;sizeof (char));} str++; //Jump knife Next substring}for  (int i = 0; i <= tmpcnt; ++i)//Output substring cout  << * (head + i); Cout << endl;} N*n   back glyphs   serpentine   arrays   output by rows// eg: n=3// 1 2 3// 8 9 4//  7 6 5void snakearr (int n) {int* arr = new int[n*n]; // Storage array int type = 0;//write array type  0  left-to-right  1  top-down  2  from right to left  3 up int   num = 1;//input Array value   1---n*nint  gap = 0;//output span per line  int  i = 0, j = 0;//array subscript while  (num <= n*n)//num add to N*n description   full fill   End Loop {if  (type == 0)//left-to-right   write array {for  (int index = i +  Gap; i < n - gap&&num <= n*n; ++i) arr[i+j*n] =  num++;++type;//Rampage Write Array completion   Type +1  variable vertical lines ++j;//j+1 to the next line}if  (type ==1)//from top to bottom {--i;//i back to   outermost for  (int index = j + gap; &NBSP;J&LT;&NBSP;N&NBSP;-&NBSP;GAP&AMP;&AMP;NUM&NBSP;&LT;=&NBSP;N*N;&NBSP;++J) arr[i + j*n] =  num++;++type;--i; //i back to outermost-1}if  (TYPE&NBSP;==&NBSP;2)//right-to-left {--j;//j fallback to not write outermost for  (int  index = i - gap; i >= gap&&num <= n*n; - -i) arr[i + j*n] = num++;++type;++i;//i to the left without traversing the outermost}if  (type == 3)//from the lower lane {--j;// J fallback to the outermost for  not written (int index = n - gap; j > gap&&num  &LT;=&NBSP;N*N;&NBSP;--J) arr[i + j*n] = num++;++j;//j back to the outermost ++i;//i not written to the left without traversing the outermost type =  0;//Write direction   type   Reset 1}gap++;//Four directions   write complete    span +1}for  (int index =  0; index < n*n; ++index) cout << arr[index] <<  '   '; Cout <<&nbSp;endl;delete arr;} In a string   delete extra words specifier   find the longest   palindrome string length//eg:abbedbba   result  abbebba  //logical large   Very complex   seems to have high-end algorithm Int plalindromenum (char* str,int start,int end) {  if  (* STR&NBSP;==&NBSP;0)//judgment   pointer validity return 0;int count = 0;//the number of characters in the current string int bigcnt  = 1;//maximum palindrome string character number int right = end;//critical value while  (start<end&&count!=2)// If you find a match for the last current character   or   string to tail   jump out loop  {//start  and end are the two identical characters of the current substring for  (int i =  start + 1; i <=right; ++i) {if  (Str[i] == str[start]) {End  = i;count = 2;//has the same item  count as 2}}if  (count != 2)//No same item   skip to next character + + Start;} if  (END&NBSP;-&NBSP;START&GT;2)//recursive   Find a pair of other identical characters in the same number Count += plalindromenum (str,  start + 1, end - 1);else if  (end - start == 2)//When End-start The 2  description contains several separate characters  count++{++count;} if  (right -end>1)//If the threshold value-end>1 the description followed by the character Traversal  {if  (Bigcnt < count)// The count  will be saved    placed 0bigcnt = count;count = 0;count += plalindromenum ( Str, end+1 , right);//start with a substring to find the matching character}if  (Bigcnt < count) bigcnt =  count;//always keep bigcnt as the number of characters of the longest palindrome string return bigcnt;} Implementation of N-bit gray code with recursion   //so far not to understand////char* gray (int n)//{//char *arry =  (char*) malloc (sizeof (char) * (int) pow (2, n) +1);//if  (n =1)//{//arry[0] =  ' 0 ';//arry[1] =   ' 1 ';//arry[2] =0 ;//return arry ;//}////char * tmp=gray (n - 1); /int j =  (int) pow (2, n)  - 2;//for  (int i = 0;i<j; ++ i)//{//arry[i] = tmp[i] + 0;//arry[j - 1 - i] = tmp[i]  + 1;//}//return arry;//}There is a string of beads 2 colors (red, blue), N (n is odd), how many blue beads can be determined no matter what order to wear//can be from different positions of two blue beads//truncated to 2 strings, and get a bunch of numbers for (n+1)/2 (not including these two Blue beads)//heavy in logical understanding   Code Simple Void pearl_num (int n) {if  (n < 3 | |  n % 2 != 1) return;int head = 0,end=0,count=0,num=2;for  (int  i = 1; i <n-2; ) {end = i;if  (end ==  (n +  1) { / 2) {printf ("%d\n",  num); return;} if  (end - head - 1)  ==  (n + 1)  / 2 | |   (n - 1 - end)  ==  (n + 1)  / 2) {i++;} else if  (count > 0) {count--;i++;} else{num++;count=num-2;i = 1;}} printf ("No answer! \ n ");} Numbers of man reading  0<=num<1000  for example   input  101  output yibailingyi void test () {int  num = 0;char* a[12] = {  "Ling",  "yi",  "er",  "san",  "Si",   "WU",  "Liu",  "Qi",   " Ba ", " Jiu ", " Shi ", " Bai " };while  (1) {Int arr[3] = { 0 };char * pinyin =  (char*) malloc (sizeof (char) *20); scanf ("%d",  &num); int tmp1 =  num;for  (int i = 2; i >= 0; --i) {arr[i] = num  % 10;num /= 10;} char* tmp = pinyin;for  (int i = 0; i<3; ++i) {if  (TMP1  == 0) {strcpy (pinyin,  "Ling\0");p Inyin += 4;break;} if  (i == 0) {if  (arr[i] != 0) {strcpy (Pinyin, a[arr[i]]);p inyin +=  (int) strlen (a[arr[i]); strcpy (pinyin, a[11]);p inyin += 3;}} else if  (i == 1) {if  (arr[i] == 0) {if  (arr[i - 1] ==  0) continue;if  (arr[i + 1] == 0) break;strcpy (pinyin, a[0]);p inyin +=  4;} else{strcpy (Pinyin, a[arr[i]);p inyin +=  (int) strlen (a[arr[i]); strcpy (pinyin, a[10]);p inyin += 3;}} else{if  (arr[i] != 0) {strcpy (pinyin, a[arr[i]);p inyin +=  (int) strlen (A[arr[i]]) ;}} free (pinyin);} *pinyin = 0;printf ("%s\n",  tmp);p inyin = tmp;}} Xiao Ming outfit equipment according to from left to right    if the item is too large   (greater than the maximum capacity of the backpack) directly discarded, if the backpack can not fit   //to replace a backpack, the previous backpack no longer use, how many items can be loaded Xiao Ming? n  Items  t  Backpack weight   m  backpack number//a[0]--a[n] Given item quality//void package () {int m, t &NBSP;,N;INT&NBSP;A[20]&NBSP;=&NBSP;{&NBSP;0&NBSP;};INT&NBSP;I&NBSP;=&NBSP;0;SCANF ("%d%d%d", &n,&t, &m);while  (i < n) {scanf ("%d",  &a[i++]);} int count = 0, tmp = t;for  (i=0; m>0 && n  > 0;) {if  (a[i] > tmp) {i++;n++;} else if  (t-a[i]>=0) {t -= a[i++];++count;--n;} else if  (m>1) {--m;t = tmp;t -= a[i++];++count;--n;    } Else{++i;--n;}} printf ("%d\n", count);} String   Press space   Flip   Simultaneous   case reversal   Example: "This is a str"->  "str a  Is this "//default given string contains only letters and spaces Void reverse_str_a (CHAR&NBSP;*&NBSP;SRC) {char* str =  (char*) malloc (sizeof (char) * (strlen (SRC)));//Copy the given string to the strcpy (STR,SRC) in an array of uppercase characters; char* tmp = str;int i  = 0;char* out =  (char*) malloc (sizeof (char) * (strlen (str)  ));//Generate stored processed character array char*  output = out;while  (* (TMP))//Calculate the number of spaces {if  (*tmp++ ==  '   ' &&*tmp!= 0)//' no calculation before ++i;} int count = i;int *a =  (int *) malloc (sizeof (int) *i);//Generate a record space subscript array tmp =  str;i = 0;int j = 0;while  (*TMP)//The space subscript is recorded and replaced with ' ' for easy copy {if  (*tmp = =  '   ')//{*tmp = 0;if  (* (tmp + 1) == 0)//' + ' before the space directly processed    Replace the first character of the output array with a space   point to the next {*output =  '   '; output++;break;} A[j++] = i;} ++tmp;++i;} tmp = str;for  (&NBSP;J&NBSP;=&NBSP;COUNT-1;&NBSP;J&GT;=0;&NBSP;--J)//writes the split string from the back to the output string and the respective \ 0 Replace with space {if  (a[0] == 0&&j==0) break;strcpy (output,tmp+a[j]+1); Output += strlen ( TMP&NBSP;+&NBSP;A[J]+1); * (output++)  =  '   ';} if  (a[0] == 0)//Processing start space tmp++;strcpy (output, tmp );//Copy the first string to the tail output +=  Strlen (tmp );if  (a[0] == 0)//move the opening space to the end {output[0]=  '   '; output[1] = 0;} i = 0;output = out;while  (* (output+i))//Replace Case {if  (* (output + i)  >=  ' A ' &&* (output + i)  <=  ' z ') * (output + i)  -=  32;else if  (* (output + i)  >=  ' A ' &&* (output + i)   <=  ' Z ') * (Output + i)  += 32;++i;} cout << out << "***********" << endl;//output free (out); FREE (output); free (a);} Int main () {cout<<cownum () <<endl;cout << cownum1 ()  << endl ; Allcolorbead ("11231324566457711234567111",  7); Allcolorbead ("ASDCCAACSDFDGFWERERDFG", 8); Snakearr (5);char* a =  "Vvaabbcdcbbaaxaabbcdcbbaax"; Cout << plalindromenum (A,  0, strlen (a))  << endl;/*pearl_num (101);p ackage (); */reverse_str_a ("         This   w W w  Is str  ");/* Test (); *//*char * aa; aa=gray ( cout << aa << endl;*/); return 0;}

C + + Some pen questions