Chapter 18: explore the wonders of the mean in the 2-bit a~b system, specify the 3x+1 conversion process of integers, specify an interval to verify the angular Valley conjecture, explore the number of

//2 position a~b Exploration of the spectacle of the mean bit in hexadecimal 
int main()
{
    
	int a, b, c, d, e, g, i, j, p, t;
	long s;
	printf("  Please enter the interval of port base a,b： ");
	scanf("%d, %d", &a, &b);
	for (p = a; p <= b; p++)
		for (j = 1; j <= p - 1; j++)
		{
    
			g = j * p + j;
			s = g * g;
			d = s;
			e = d % p;
			d = d / p;
			t = 0; 
			while (d > 0)
			{
    
				c = d % p;
				d = d / p;
				if (c != e)
				{
    
					t = 1;
					break;
				}
			}
				
						if (t == 0)
						{
    
							printf("  stay %d  In base ： ", p);
							for (i = 1; i <= 2; i++)
								if (j < 10)
									printf("%ld", j);
								else
									printf("[%ld]", j);
							printf(" ^2=");
							while (s > 0)
							{
    
								c = s % p;
								s = s / p;
								if (c < 10)
									printf("%ld", c);
								else
									printf("[%ld] ", c);
							}
								printf("\n");
						}
		}
	printf(" stay %d-%d We have explored the above wonders of the mean bit \n", a, b);
	return 0;

}

result ：

// Specify the  3x+1  The transformation process 
int main()
{
    
	long int m, n, c;
	printf("  please enter an integer  m： ");
	scanf(" %ld", &m);
	n = 0;
	c = m;
	printf(" %ld", m); 
	while (c != 1)
	{
    
		if (c % 2 == 1)
		{
    
			c = 3 * c + 1;
			n++; // Odd hour , multiply 3 After add 1
			printf("->%ld", c);
		}
		else
		{
    
			c = c / 2;
			n++; //  Even number , Divide 2 
			printf("->%ld", c);
		}
		if (n % 10 == 0)
			printf("\n");
		if (n > 10000) break;
	}
	if (n > 10000)
		printf("\n Exceed  10000  Step has not completed the conversion .");
	else printf("\n Together with  %ld Step complete the conversion .\n",n);


	return 0;
}

result ：

// Specify an interval to verify the angular Valley conjecture 
int main()
{
    

	 long a, b, c, m,m1, m2,n,max,min;
	 printf(" Please enter the interval limit  a, b: ");
	 scanf ("%ld,%ld", &a, &b) ;
	 max=0;
	 min=100;
	 for (m=a;m<=b;m++) 
	 {
     
		 n=0; c=m; 
		 while (c != 1)
		 {
    
			 if (c % 2 == 1) //  Odd hour , multiply 3 After add 1

			 {
    
				 c = 3 * c + 1;
				 n++;
			 }
			 else
			 {
    
				 c = c / 2;
				 n++; // Even number , Divide 2

			 }
		 }
			 if (n > max) {
    
				 max = n;
				 m1 = m;
			 } // Find the maximum value of the conversion step 
			 if (n < min) {
    
				 min = n; m2 = m;
			 } // Find the minimum value of the conversion step  
	 }
		 printf("  When m=%ld  The maximum number of conversion steps is , Together with %ld Step through .\n",m1,max);
		 printf("  When m=%1d The number of conversion steps is the least , Together with %1d Step through .\n", m2, min);


	return 0;
}

result ：

//  verification 3 Number of bit black holes 
int  main()
{
    
	int i, j, n, n1, m, max;
	int sub(int i);
	printf(" Verify any 3 digit “ Rearrange the difference ” operation , To  495  or 0. \n");
	for (max = 0, n = 100; n <= 999; n++)
	{
    
		i = n;
		m = 0;
		while (i != 495 && i != 0) //m Statistics are converted to  495  The number of times  //  Conditional discrimination  
		{
    
			i = sub(i);
			m++;
		}
		if (m > max)
		{
    
			max = m;
			n1 = n;
			j = i; // Maximum number of conversions  max
		}
	}


	if (j == 195)
		printf(" When n=%d when , At most  % d  Secondary arrival  495: \n", n1, max);
	if (j == 0)
		printf(" When  n = %d  when , At most xd  Secondary arrival  0: \n", n1, max);
	printf(" %d", n1);
	i = n1;
	while (i != 495 && i != 0) 
	{
    
		i = sub(i);
	printf("->%d",i);
	}
	printf(".\n");
}
int sub(int i)
{
    
	int p, k, j, h, max, min, a[4]; 
		a[1] = i / 100;
	a[2] = (i / 10) % 10;
	a[3] = i % 10;
	for (k = 1; k <= 2; k++)
		for (j = k + 1; j <= 3; j++)
			if (a[k] < a[j])
			{
    
				h = a[k];
				a[k] = a[j];
				a[j] = h;
			}

	max = a[1] * 100 + a[2] * 10 + a[3];
	min = a[3] * 100 + a[2] * 10 + a[1];

	p = max - min;
	if (p < 100 && p / 10 == p % 10) p = 0;
	return (p);
}

result ：

// Seek 4 Number of black holes  
int main()
{
    

	int i, j, m, n, n1, max;
	int sub(int i);
	printf("  all 4 The number of digits is “ Rearrange the parameters ” operation , can 6194 or 0\n");
	for (max = 0, n = 1000; n <= 9999; n++)
	{
    
		i = n;
		m = 0;
		while (i != 6174 && i != 0)
		{
    
			i = sub(i);
			m++;
		}
		if (m > max)
		{
    
			max = m;
			n1 = n;
			j = i;
		}
	}
	if (j == 6174)
		printf(" \n %d when , At most %d Secondary arrival  6174: \n", n1, max); 
	if(j == 0)
		printf(" \n %d when , At most %d Secondary arrival  0: \n", n1, max);
	printf(" %d",n1);
	i = n1;

	while (i != 6174 && i != 0)
	{
    
		i = sub(i);
		printf("->%d", i);
	}
	printf(".\n");
}

int sub(int i)
{
    
	int p, k, j, h, max, min, a[5];
	a[1] = i / 1000;
	a[2] = (i / 100) % 10; //  hold 1 digit 1 Decompose into 4 A digital  
	a[3] = (i % 100) / 10;
	a[4] = i % 10;
	for (k = 1; k <= 3; k++)
		for(j = k+1;j <=4;j++)
		if (a[k] < a[j])
		{
    
			h = a[k];
			a[k] = a[j];
			a[j] = h;
		}
	max = a[1] * 1000 + a[2] * 100 + a[3] * 10 + a[4];
	min = a[4] * 1000 + a[3] * 100 + a[2] * 10 + a[1];
	p = max - min;
	if (p < 1000 && p / 100 == (p / 10) % 10 && p / 100 == p % 10)
		p = 0;
	return (p);
}

result ：