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La touche de début ou touche d'origine, également désignée par l'anglicisme « Home », est une touche de clavier d'ordinateur qui permet de retourner en début de ligne ou tout en haut d'une page. Elle est souvent représentée par une flèche dirigée vers le coin supérieur gauche, par exemple ↖ ou ◤, mais peut également l'être par l’abréviation Orig. Sur certains claviers, l'abréviation pourra être Pos1. Elle fonctionne de façon opposée à la touche de fin et est souvent voisine des touches de défilement et Inser / Suppr.
Sur Microsoft Windows et Linux, dans les documents éditables tels que ceux ouverts dans un traitement de texte, cette touche ramène le curseur au début de la ligne, et associée à Ctrl, au début de la page ; tandis que dans les documents non éditables, elle permet de retourner au début de la page quoi qu'il arrive.
Dans les applications Mac OS X, elle fait défiler le document jusqu'au début, mais sans modifier la position du curseur.
Dans certains navigateurs web, elle est utilisée dans une combinaison de touche permettant de recharger la page d'accueil.
What else can I help you with?
What is pos1 button on computer keyboard?
it's the HOME key (above arrows)
What is recursive algorithm and write a recursive algorithm to reverse a string also trace it for any string data?
A recursive algorithm is an algorithm which calls itself with "smaller (or simpler)" input values, and which obtains the result for the current input by applying simple operations to the returned value for the smaller (or simpler) input. Heres a recursive algorithm to reverse a string char *rev(char str[],int pos1,int pos2) { if(pos1<pos2) { char temp=str[pos1]; str[pos1]=str[pos2]; str[pos2]=temp; return rev(str,pos1+1,pos2-1); } return str; } You can call this function like this char *r=rev("reverse it",0,9);
How do you reset service massage for Volvo s 80?
Ignition to Pos1 Push and hold in odometer reset button Ig switch to Pos2 Wait until service light blinks, then release odometer button
How do you write a c plus plus program to calculate the number of days between two dates?
#include<iostream> #include<vector> #include<array> #include<sstream> struct date { size_t day; size_t month; size_t year; bool is_leap_year () const { if (year%4) return false; // If not a multiple of 4 then common year. if (year%100) return true; // If a multiple of 4 but not of 100 then leap year. if (year%400) return false; // If a multiple of 4 and 100 but not of 400 then common year. return true; // If a multiple of 4, 100 and 400 then leap year. } bool is_valid () const { if (!day 31<day) return false; if (!month 12<month) return false; if (year<1745) return false; if (30<day && (month==4 month==6 month==9 month==11)) return false; if (month==2 && 28<day && !is_leap_year()) return false; return true; } size_t days_between_years (const date& other) const { size_t begin = year < other.year ? year : other.year; size_t end = year < other.year ? other.year : year; size_t days = 0; while (begin < --end) days += date{1,1,end}.is_leap_year() ? 366 : 365; return days; }; size_t day_of_year () const { size_t days = day; size_t m = month; while (--m) days += date{1,m,year}.days_in_month(); return days; } size_t days_in_month () const { switch (month) { case 2: return is_leap_year () ? 28 : 29; case 4: case 6: case 9: case 11: return 30; } return 31; } }; bool operator input.npos pos1==pos2) { std::cerr << "Invalid date. Please use '/' separator.\n"; continue; } std::vector<std::string> split = split_string (input, '/'); if (split.size() != 3) { std::cerr << "Invalid date. Please use 'dd/mm/yyyy' format.\n"; continue; } std::array<size_t, 3> nums = {1,1,1980}; for (size_t i=0; i<3; ++i) { std::stringstream ss; ss << split[i]; ss >> nums[i]; } d.day = nums[0]; d.month = nums[1]; d.year = nums[2]; if (!d.is_valid()) { std::cerr << "Invalid date.\n"; continue; } return d; } } int main() { while (true) { date d1 = enter_date("Enter 1st date: "); date d2 = enter_date("Enter 2nd date: "); if (!(d1<d2)) { std::cerr << "Error: first date must come before second date.\n"; continue; } size_t days = d2 - d1; std::cout << "There are " << days << " days between the two dates\n"; break; } }
C program to for traveling salesman problem?
#include<conio.h> #include<stdio.h> #include<stdlib.h> #define N 50 //Travelling sales man problem , travelling between 6 cities. //initializing the path lengths between cities and the paths to be included //in population void initialize(int pathlen[][6],int path[][6]) { int i,j,k; //obtaining pathlengths for(i=0;i<6;i++) { for(j=0;j<6;j++) { if(j<i) //path length from a to b will be same as b to a { pathlen[i][j]=pathlen[j][i]; } if(j==i) // path length from a to a will be 0 { pathlen[i][j]=0; } if(j>i) // rest initialized { do{ pathlen[i][j]= random(20); }while(!pathlen[i][j]); } } } // display the path lengths printf("\n\tThe PATH LENGTHS ARE: \n" ); for(i=0;i<6;i++) { for(j=0;j<6;j++) { printf(" %5d ",pathlen[i][j]); } printf("\n\n"); } // generating the population for(i=0;i<20;i++) { for(j=0;j<6;j++) { path[i][j]=random(6); for(k=j-1;k>=0;k--) { if(path[i][j]==path[i][k]) //checking to avoid repeatition { path[i][j] = random(6); k=j; } } } } } // evaluating the fitness function or total distance void evaluate(int pathlen[6][6],int path[20][6],int fx[20]) { int sum =0,i,j,a,b; //obtaing the sum of the path taken for(i=0;i<20;i++) { sum=0; for(j=0;j<5;j++) { a=path[i][j]; b=path[i][j+1]; sum=sum+pathlen[a][b]; } fx[i]=sum; } //display the paths generated printf("\n"); printf("\n\tPATH \t\tf(x) \n\n"); for(i=0;i<20;i++) { printf("\t"); for(j=0;j<6;j++) { printf(" %d",path[i][j]); } printf("\t%d",fx[i]); printf("\n"); } } //selecting the two points for cross over and then performing partial Crossover void selection(int fx[20],int pos[2],int posmax[2]) { int min1=fx[0],min2=fx[0],i,max1=fx[0],max2=fx[0]; pos[0]=0; pos[1]=0; posmax[0]=0; posmax[1]=0; //calculating the minimum postion for(i=1;i<20;i++) { if(fx[i]<min1) { min1=fx[i]; pos[0]=i; } } //calaculating the second minimum position for(i=1;i<20;i++) { if(fx[i]<min2&&i!=pos[0]) { min2=fx[i]; pos[1]=i; } } //calculating the max position for(i=1;i<20;i++) { if(fx[i]>max1) { max1=fx[i]; posmax[0]=i; } } //calculating the second max position for(i=1;i<20;i++) { if(fx[i]>max2&&i!=posmax[0]) { max2=fx[i]; posmax[1]=i; } } printf("\n\tFIRST MINIMUM=%4d \tPOSITION=%4d\n\tSECOND MINIMUN=%4d \tPOSITION=%4d\n\tFIRST MAXIMUM=%4d \tPOSITION=%4d\n\tSECOND MAXIMUM=%4d \tPOSITION=%4d\n",min1,pos[0],min2,pos[1],max1,posmax[0],max2,posmax[1]); } //PERFORMING PARTIAL CROSSOVER void crossover(int pos[2],int path[][6],int child[2][6]) { int crosspt1,crosspt2,j,i,temp,temp1[2][6],temp2; //TAKING 2 CROSS POINTS do { crosspt1=random(5); }while(crosspt1>2) ; do { crosspt2=random(5); }while(crosspt2<=3); clrscr(); printf("\n\n\t The CROSSOVER POINTS ARE : %d , %d ",crosspt1,crosspt2); printf("\n\n\tTHE PATHS FOR CROSSOVER ARE"); printf("\n\n\t\t"); for(j=0;j<6;j++) { child[0][j]=path[pos[0]][j]; printf(" %d",child[0][j]); } printf("\n\t\t"); for(j=0;j<6;j++) { child[1][j]=path[pos[1]][j]; printf(" %d",child[1][j]); } int cnt=0; //swapping the paths between two crosspoints for(j=crosspt1+1;j<=crosspt2;j++) { temp1[1][cnt]=child[0][j]; temp1[0][cnt]=child[1][j]; temp=child[0][j]; child[0][j]=child[1][j]; child[1][j]=temp; cnt++; } //performing partial crossover int k,m; for(m=0;m<2;m++) { for(i=0;i<crosspt1+1;i++) //taking the path before crosspoint { for(j=0;j<cnt;j++) //comparing the path within crossover point { if(child[m][i]==temp1[m][j]) //if found then { if(m==0) //for child 1 { temp2=temp1[1][j]; //take the path from child2 crossover for(k=0;k<6;k++) { if(child[m][k]==temp2) //if still the path repeats then repeat the process again { temp2=child[1][k]; k=0; } } child[m][i]=temp2; //finally putting the value in child } else //for child 2 { temp2=temp1[0][j]; for(k=0;k<6;k++) { if(child[m][k]==temp2) {temp2=child[0][k]; k=0; } } child[m][i]=temp2; } } } } } for(m=0;m<2;m++) { for(i=crosspt2+1;i<6;i++) //now chehcking the path after the second cross point { for(j=0;j<cnt;j++) //comparing the path within crossover point { if(child[m][i]==temp1[m][j]) //if found then { if(m==0) //for child 1 { temp2=temp1[1][j]; //take the path from child2 crossove for(k=0;k<6;k++) { if(child[m][k]==temp2) //if still the path repeats then repeat the process again {temp2=child[1][k]; k=0; } } child[m][i]=temp2; //finally assigning the value } else //for child 2 { temp2=temp1[0][j]; for(k=0;k<cnt;k++) { if(child[m][k]==temp2) {temp2=child[0][k]; k=0; } } child[m][i]=temp2; } } } } } //display AfTER CROSSOVER printf("\n\tAFTER CROSSOVER\n\t\t"); for(j=0;j<6;j++) { printf(" %d",child[0][j]); } printf("\n\t\t"); for(j=0;j<6;j++) { printf(" %d",child[1][j]); } } //insering the paths in population removing those having maximum populaiton void insert(int child[2][6],int posmax[2],int path[20][6]) { for(int j=0;j<6;j++) { path[posmax[0]][j]=child[0][j]; path[posmax[1]][j]=child[1][j]; } } // performing mutation void mutation(int child[2][6]) { int sel=random(2); int pos1=random(6); int pos2=random(6); int temp=child[sel][pos1]; child[sel][pos1]=child[sel][pos2]; child[sel][pos2]=temp; } void main() { clrscr(); randomize(); int pathlen[6][6],path[20][6],fx[20],pos[2],posmax[2],child[2][6]; printf("\n\t\t\t TRAVELLING SALESMAN PROBLEM "); printf("\n\t\t\t_____________________________"); printf("\n\n\n\t\tTHE TRAVELLING SALES MAN PROBLEM DEALS WITH THE FACT"); printf("\n\n\t\tTHAT A SALESMAN TRAVELS BETWEEN CITIES TAKING THE PATH"); printf("\n\n\t\tTHAT IS OF MINIMUN DISTANCE."); printf("\n\n\n\t\tTO OBTAIN THE MINIMUM DISTANCE WE USE GENETIC ALGO"); printf("\n\n\t\tWHERE WE TAKE AN INITIAL POPULATION OF 20 PATHS AND "); printf("\n\n\t\tAND THEN THROUGH FITNESS FUNCTION WE OBTAIN THE PATHS "); printf("\n\n\t\tWITH MINIMUN DISTANCE AND REPLACE THEM WITH THE CHILDS "); printf("\n\n\t\tAFTER PARTIAL CROSSOVER WITH MAXIMUM DISTANCE."); getch(); clrscr(); initialize(pathlen,path); evaluate(pathlen,path,fx); getch(); selection(fx,pos,posmax); crossover(pos,path,child); mutation(child); getch(); insert(child,posmax,path); for(int i=1;i<N;i++) { evaluate(pathlen,path,fx); selection(fx,pos,posmax); crossover(pos,path,child); mutation(child); insert(child,posmax,path); } evaluate(pathlen,path,fx); selection(fx,pos,posmax); crossover(pos,path,child); insert(child,posmax,path); evaluate(pathlen,path,fx); getch(); }
