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Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
P1 or parental
P1-e is an expression, not a formula.
In genetics, in a pure-breeding population, the parental generation is the P1 generation. The off-spring of the P1 Generation is called the F1 Generation
If the old population is P1, the new population is P2, and the growth rate is G, G = (P2 - P1) ÷ P1 x 100%
P1: tt F2: tt
#include#include#define n 100void del_char(char str1[n],char str2[]){char str[n],*p1,*q,i,j;p1=str1;q=str;*q=*p1;i=0;while(*p1!='\0'){{if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;}else{q++;}}else{q++;}}else{p1++;q++;}*q=*p1;}i=0;}printf("%s\n",str);}main(){char str1[n],str2[]={"to "};clrscr();printf("please enter a text and includ 'to':\n");gets(str1);printf("remove 'to',remaining :\n");del_char(str1,str2);getch();return 0;}Output:please enter a text and includ 'the':Write a c program to simulate the calculator using functionremove 'the',remaining :Write a c program to simulate calculator using function
P1 Material is the name given to Carbon Manganese Steel in the engineering industry
ed=(q1-q2)/q1/(p1-p2)/p1
P1" is the highest short-term rating category for Moody's Investor Service. P1 rating are considered to be of high credit quality
(p1/v1) = (p2/v2)For Apex (P1 N1)= (P2N2 )
#include<stdio.h> #include<conio.h> int main() { char p[10][5],temp[5]; int i,j,pt[10],wt[10],totwt=0,pr[10],temp1,n; float avgwt; clrscr(); printf("enter no of processes:"); scanf("%d",&n); for(i=0;i<n;i++) { printf("enter process%d name:",i+1); scanf("%s",&p[i]); printf("enter process time:"); scanf("%d",&pt[i]); printf("enter priority:"); scanf("%d",&pr[i]); } for(i=0;i<n-1;i++) { for(j=i+1;j<n;j++) { if(pr[i]>pr[j]) { temp1=pr[i]; pr[i]=pr[j]; pr[j]=temp1; temp1=pt[i]; pt[i]=pt[j]; pt[j]=temp1; strcpy(temp,p[i]); strcpy(p[i],p[j]); strcpy(p[j],temp); } } } wt[0]=0; for(i=1;i<n;i++) { wt[i]=wt[i-1]+et[i-1]; totwt=totwt+wt[i]; } avgwt=(float)totwt/n; printf("p_name\t p_time\t priority\t w_time\n"); for(i=0;i<n;i++) { printf(" %s\t %d\t %d\t %d\n" ,p[i],pt[i],pr[i],wt[i]); } printf("total waiting time=%d\n avg waiting time=%f",tot,avg); getch(); }