예제 #1
0
파일: tree.c 프로젝트: abhisg/allcodes
int main()
{
	long long int n,co,i,min=0;
	scanf("%lld",&n);
	if(n==2)
	{
		printf("%d\n",0);
	}
	else
	{
		long long int *coor=(long long int *)malloc(n*sizeof(long long int));
		scanf("%lld",coor);
		scanf("%lld",coor+1);
		scanf("%lld",coor+2);
		long long int gcd=hcf(coor[1]-coor[0],coor[2]-coor[1]);
		for(i=3;i<n;i++)
		{
			scanf("%lld",coor+i);
			gcd=hcf(gcd,coor[i]-coor[i-1]);
		}
		for(i=1;i<n;i++)
		{
			min+=(coor[i]-coor[i-1])/gcd-1;
		}
		printf("%lld\n",min);
		return 0;
	}
}
void main()
{
	clrscr();
	int tes,tot,hc;
	cin>>tes;
	for(int a=1;a<=tes;a++)
	{
		cin>>tot;
		int *ar=new int[tot];
		for(int i=0;i<tot;i++)
		{
			cin>>ar[i];
		}
		hc=hcf(ar[0],ar[1]);
		for(int j=2;j<tot;j++)
		{
			hc=hcf(hc,ar[j]);
		}
		for(int k=0;k<tot;k++)
		{
			cout<<ar[k]/hc<<" ";
		}
		cout<<endl;
	}
	getch();
}
예제 #3
0
파일: hcf.cpp 프로젝트: malavdk/codes
main( ) {
int a, b, result;

printf( "\n enter two numbers: ");
scanf( "%d%d", &a, &b);
	     result = hcf(a, b);
	     printf( "\n The hcf is %d: ", result);
}
예제 #4
0
void main()
{
	int x, y, h;
	printf("\nEnter 2 numbers to find hcf:");
	scanf("%d%d", &x, &y);
	h=hcf(x, y);
	printf("\n%d", h);
}
예제 #5
0
int hcf(int a, int b)
{
	if(a%b==0)
	return b;
	if(a%b==1)
	return 1;
	return hcf(b,(a%b));
}
예제 #6
0
int main()
 {void hcf(int,int);
  void lcd(int,int); 
  int u,v;
  scanf("%d,%d",&u,&v);
  hcf(u,v);
  lcd(u,v);
  printf("H.C.F=%d\n",Hcf);
  printf("L.C.D=%d\n",Lcd);
  return 0;
 }
예제 #7
0
int main()
{
    int a, b, result;
 
    printf("Enter the two numbers to find their HCF: ");
    scanf("%d%d", &a, &b);
    result = hcf(a, b);
    printf("The HCF of %d and %d is %d.\n", a, b, result);
 
    return 0;
}
int main()
{int hcf(int,int);
 int lcd(int,int,int);
 int u,v,h,l;
 scanf("%d,%d",&u,&v);
 h=hcf(u,v);
 printf("H.C.F=%d\n",h);
 l=lcd(u,v,h);
 printf("L.C.D=%d\n",l);
 return 0;
}
int main()
{
	void hcf(int,int);
	void lcd(int,int);
	int u,v;
	scanf("%d,%d",&u,&v);
	hcf(u,v);
	lcd(u,v);
	printf("最大公约数是%d\n",Hcf);
	printf("最小公倍数是%d\n",Lcd);
	return 0;
}
예제 #10
0
파일: recipe.c 프로젝트: Garvit/code-backup
int main()
{
    int t,n,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int gcd,j;
        scanf("%d",&arr[0]);
        gcd=arr[0];
        for(j=1;j<n;j++)
        {
            scanf("%d",&arr[j]);
            gcd=hcf(gcd,arr[j]);
        }
        for(j=0;j<n;j++)
        {
            printf("%d ",(arr[j]/gcd));
        }
        printf("\n");

    }
    return 0;
}
예제 #11
0
int main()
{
	int t,n,i; 
	long long int x,a,b,end;

	scanf("%d",&t);
	x=0;
	while(t--)
	{
		scanf("%d",&n);
		//printf("%lld",x);

		for(i=0;i<n;i++)
		{
			scanf("%lld",&x);
			if(i==0) end=x;
			else end=hcf(end,x);
			//printf(">>%lld",end);

		}
		printf("%lld\n",end);
	}
	return 0;
}
예제 #12
0
int main() {
    int a,b;
    printf(" Enter Two Numbers : ");
    scanf("%d%d",&a,&b);
    printf(" GCD = %d \n HCF = %d\n",gcd(a,b),hcf(a,b));
}
예제 #13
0
int lcm(int a,int b)
{
    return a*(b/hcf(a,b));
}
예제 #14
0
long long int hcf(long long int a,long long int b){
    if(b>a) return hcf(b,a);
    return b==0?a:hcf(b,a%b);
}