!!! "! Backtrackin # g % $ 14

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Ursula Harrington
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1 backtracking

2 " Backtracking# % $

3 Structure & ' ' ( ' ( % $( "

4 struct <name> { <type1> <member name1>;... <typen> <member namen>; ; #define NAME_SIZE 20 #define ID_SIZE 10 struct address { char city[name_size]; char street[name_size]; int house_number; ; struct date { int day, month, year; ; struct employee { char name[name_size]; char id[id_size]; double salary; struct date birth_date; struct address addr; ; )

5 struct date { int day, month, year; ; struct dateday, month, year % members$ +++ struct dated & ddayd.day, struct date d; ++ & emp.addr.city struct_name.member_name = x struct_namemember_name 'x *

int main() { struct date d1 = {2, 5, 1975; struct address home = {"Los-Angeles", Kipling", 711; struct employee beckham; strcpy(beckham.name, "David Beckham"); strcpy(beckham.

6 int main() { struct date d1 = {2, 5, 1975; struct address home = {"Los-Angeles", Kipling", 711; struct employee beckham; strcpy(beckham.name, "David Beckham"); strcpy(beckham.id, " "); beckham.salary = ; beckham.birth_date = d1; beckham.addr = home;... return 0; struct employee { char name[name_size]; char id[id_size]; double salary; struct date birth_date; struct address addr; ;,

7 by-value typedef struct date date_t; date_t earlier_date(date_t d1, date_t d2) { if ( d1.year < d2.year ) return d1; if ( d1.year > d2.year ) return d2; if ( d1.month < d2.month ) return d1; if ( d1.month > d2.month ) return d2; return (d1.day < d2.day)? d1 : d2; earlier_date() date_t earlier_date()' -

8 by-address ' +->+ % (*p).m#p->m$ +.++->+ boolean before(date_t* pd1, date_t* pd2) { if((pd1->year < pd2->year) ((pd1->year == pd2->year) && (pd1->month < pd2->month))...))) return TRUE; return FALSE;.

9 struct employee beckham, employees[num_of_empl]; double sum = 0; int counter = 0;... for(i = 0; i < NUM_OF_EMPL; i++) { if(before(&employees[i].date, &beckham.date)) { sum += (employees + i)->salary; counter ++; return sum / counter; /

10 #define NAME_SIZE 20 #define ID_SIZE 10 struct date { int day, month, year; ; struct address { char city[name_size]; char street[name_size]; int house_number; ; struct employee { char name[name_size]; char id[id_size]; double salary; struct date birth_date; struct address addr; ; sizeof (struct date) =? 3 * sizeof (int) = 12 sizeof (struct address) =? 2 * 20 * sizeof (char) + sizeof (int) = 44 sizeof (struct employee) = 20? * sizeof (char) + 10 *sizeof (char) + sizeof (double) + sizeof (struct date) + sizeof (struct address) = =94 0

11 % struct$+ struct ArabaritWord { char* s;// null terminated int len;// number of letters ++ '& ; +#++# % lower case$ % upper case$ #num_of_lower() ++,+sababibabi+ 00

12 % struct$+ struct ArabaritWord { char* s;// null terminated int len;// number of letters ; ++ & '& #num_of_lower() ++ 0' islower % ctype.h#$lower-case 0

13 " int num_of_lower (struct ArabaritWord word) { int bot = 0, top = word.len-1, mid; if (islower(word.s[bot])) return 0; if (islower(word.s[top])) return word.len; while (top>bot){ mid = (top+bot)/2; if (islower(word.s[mid])) bot = mid+1; else top = mid; return top; 0"

14 $ # #define NUMWORDS 12 #define WORD_MAX_LEN 10 struct word{ unsigned int pos; int main(){ unsigned int len; ; char s[word_max_len]; typedef struct word Word; int w; char* cname= IntroWetoLikeCS_ ; Word words[numwords]={{13,2,{23,1,{17,4,{23,1,{8,5, {32,1,{15,2,{23,1,{21,2,{23,1,{3,5,{2,1; for (w=0; w<numwords; w++){ strncpy(s,cname+words[w].pos,words[w].len); s[words[w].len]= \0 ; printf( %s,s); printf( \n ); return 0; 0)

15 723411IntroWetoLikeCS_ int main(){ RUN char s[word_max_len]; int w; char* cname= IntroWetoLikeCS_ ; Word words[numwords]={{13,2,{23,1,{17,4,{23,1,{8,5, {23,1,{15,2,{23,1,{21,2,{23,1,{3,5,{2,1; for (w=0; w<numwords; w++){ strncpy(s,cname+words[w].pos,words[w].len); s[words[w].len]= \0 ; printf( %s,s); Printed Out: printf( \n ); return 0; We_ Like_ Intro_ to_ CS_ *

16 $ % #include<stdio.h> struct vertex{ double x,y; char* desc; v1, v2, *v3; int main(){ v1.x = 1.0; v1.y = 0.0; v1.desc = "Laflaf"; v2.x = 2.0; v2.y = 1.0; v2.desc = "Arbol"; v3 = &v2; v1.desc = v2.desc; ++v3->y; printf("v1.x=%f\n v2.y=%f\n",v1.x,v2.y); printf("*(v1.desc+3)=%c\n",*((v1.desc)+3)); return 0; Printed Out: RUN v1.x=1.0 v2.y=2.0 *(v1.desc+3)=o 0,

17 " Backtracking# 0-

18 ? #&# & F,O,U,R,T,W + T W O / = T W O F O U R & backtracking#1 ' # 0.

19 ' [,,,,, ] $=[O,R,W,U,T,F] [0,,,,, ] [1,,,,, ] [2,,,,, ] [3,,,,, ] [4,,,,, ] [2,0,,,, ] [2,1,,,, ] [2,2,,,, ] [2,3,,,, ] [2,4,,,, ] [2,2,0,,, ] [2,2,1,,, ] [2,2,2,,, ] [2,2,3,,, ] [2,2,4,,, ] [2,2,3,0,, ] [2,2,3,1,, ] [2,2,3,2,, ] [2,2,3,3,, ] [2,2,3,4,, ] %2 $

20 ( void perm(int word[], int n, int index, int used[]) { if(index == n) { print_array(word, n); return; &Backtracking void riddle(int result[], int index, int used[]) { if(index == n) { if(check(result)) print_result(result,n); return; for(int i = 0; i < K; i++) { if(used[i]) continue; word[index] = i; used[i] = 1; perm(word,n,index+1,used); used[i] = 0; return; 0"++ for(int i = 0; i < 10; i++) { if (used[i]) continue; result[index] = i; used[i]=1; riddle(result,index+1,used); used[i]=0; return; % 1 $

21 T W O + T W O = F O U R & 1.(O+O)%10=R, & define r1=(o+o)/10. 2.(r1+W+W)%10=U, & define r2=(r1+w+w)/10. 3.(r2+T+T)%10=O, & define r3=(r2+t+t)/ R3=F. 0

22 ) void check(int a[]) { int r1, r2, r3; r1=(a[0]+a[0])/10; r2=(a[2]+a[2]+r1)/10; r3=(a[4]+a[4]+r2)/10; if (((a[0]+a[0])%10==a[1]) && ((r1+a[2]+a[2])%10==a[3]) && ((r2+a[4]+a[4])%10==a[0]) && (r3==a[5])) return 1; return 0; 1.(O+O)%10=R, r1=(o+o)/10. 2.(r1+W+W)%10=U, r2= (r1+w+w)/10. 3.(r2+T+T)%10=O, r3=(r2+t+t)/ R3=F. a[0] a[1] a[2] a[3] a[4] a[5] O R W U T F

23 #include<stdio.h> #include<stdlib.h> int main(){ int result[6]; int used[10]={0; RUN printf("solving \n"); printf(" T W O \n"); printf(" + T W O \n"); printf("= \n"); printf(" F O U R\n\n\n"); printf( Solutions are:\n"); riddle(result,0,used); system("pause"); return 0; void print_result(int a[]) { int i; printf("o = %d \n", a[0]); printf("r = %d \n", a[1]); printf("w = %d \n", a[2]); printf("u = %d \n", a[3]); printf("t = %d \n", a[4]); printf("f = %d \n", a[5]); printf("\n"); printf(" %2d%2d%2d \n", a[4],a[2],a[0]); printf("+ %2d%2d%2d \n", a[4],a[2],a[0]); printf("= \n"); printf(" %2d%2d%2d%2d\n\n\n", a[5],a[0],a[3],a[1]); "

24 ( void riddle(int result[], int index, int used) { if(index == n) { if(check(result)) &' print_result(result,n); return; for(int i = 0; i < 10; i++) { if (used[i]) continue; result[index] = i; used[i]=1; riddle(result,index+1,used); used[i]=0; return; void riddle(int result[], int index, int used) { if(index == n) { print_result(result,n); return; for(int i = 0; i < 10; i++) { if (used[i]) continue; if (legal(result,index,i) continue; result[index] = i; used[i]=1; riddle(result,index+1,used); used[i]=0; return; )

25 legal void legal(int a[], int index, int i) { switch (index){ RUN case 0: return 1; case 1: if ((a[0]+a[0])%10==i) return 1; break; case 2: return 1; case 3: int r1=(a[0]+a[0])/10; if ((r1+a[2]+a[2])%10==i) return 1; break; case 4: int r1=(a[0]+a[0])/10; int r2=(a[2]+a[2]+r1)/10; if ((r2+i+i)%10==a[0]) return 1; break; case 5: int r1=(a[0]+a[0])/10; int r2=(a[2]+a[2]+r1)/10; int r3=(a[4]+a[4]+r2)/10; if (r3==i) return 1; break; return 0; 1.(O+O)%10=R, r1=(o+o)/10. 2.(r1+W+W)%10=U, r2= (r1+w+w)/10. 3.(r2+T+T)%10=O, r3=(r2+t+t)/ R3=F. a[0] a[1] a[2] a[3] a[4] a[5] O R W U T F *

26 " Backtracking#,

27 ' ' % $ ) C# -

28 .

Object oriented programming

Object oriented programming Exercises 1 Version 1.0, 16 February, 2017 Table of Contents 1. Structures................................................................... 1 1.1. Date....................................................................