第二章：基于分解的求水仙花数,基于组合的求水仙花数， 兰德尔数,求[x,y]内的守形数，探求n位守形数，递推探索n位逐位整除数

//基于分解的求水仙花数程序
int main()
{
    
	int m, a, b, c;
	for (m = 100; m <= 999; m++)
	{
    
		a = m / 100;
		b = (m / 10) % 10;
		c = m % 10;
		if (m == a * a * a + b * b * b + c * c * c)
			printf(" %d", m);
	}
	return 0;
}

结果：

//基于组合的求水仙花数程序
int main()
{
    
	int m, a, b, c;
	for(a = 1; a <= 9; a++)
		for(b = 0; b <= 9; b++)
			for (c = 0; c <= 9; c++)
			{
    

				m = a * 100 + b * 10 + c;
				if (m == a * a * a + b * b * b + c * c * c)
					printf(" %d", m);
			}
			return 0;
}

结果：

//兰德尔数
int main()
{
    
	int m, n, i;
	double f, k, s, t, y;
	printf("请输入位数n (2 < n < 10):");
	scanf("%d", &n);
	m = 0;
	t = 1;
	for (i = 1; i <= n - 1; i++)
		t = t * 10;
	for (y = t + 1; y <= t * 10 - 1; y++)
	{
    
		f = y;
		for (s = 0, i = 1; i <= n; i++) // 循环分离y的n个数字k 
		{
    
			k = fmod(f, 10);
			s += pow(k, n);
		
			f = floor(f / 10); //求y的n个数字的n次幂之和s
		}
		if (y == s)
		{
    
			m++;
			printf(" %.0f", y); // 检测是否满足条件 //输出探索结果
		}
	}
	printf(" \n %d位兰德尔数共%d个", n, m);
			return 0;
}

结果：

// 求[x,y]内的守形数
int main()
{
    
	long a, b, c, k, m, s, x, y; 
	printf(" 请输入区间上下限整数x,y：");
		scanf ("%ld" , &x) ; 
		scanf("%ld",  &y);
			m=0;
	for (a = x;a <= y;a++)
	{
    
		s=a*a; //计算a的平方数s 
		b = 1;
		k=a;
		while (k > 0)
		{
    
			b = b * 10;
			k = k / 10;
		}
		c = s % b;  //为a的平方数s的尾部 
			if (a == c)
			{
    
				m++;
				printf("%ld^2 = %1d \n", a, s);
			}
	}
			printf(" 区间[%1d,%1d]中，共以上%d个守形数。", x, y, m);
	
			return 0;
}

结果：

//探求n位守形数
int main()
{
    
	int n, d, k, j, i, t, m, w, z, u, v, a[500], b[500], c[500];
	printf(" 请输入整数n：");
	scanf("%d", &n);
	for (d = 1; d <= 9; d++)
	{
    
		for (k = 1; k <= 499; k++)
		{
    
			a[k] = 0;
			b[k] = 0;
			c[k] = 0;
		}
		a[1] = d; // 给守形数个位数赋值 
		for (k = 2; k <= n; k++)
		{
    
			for (j = 0; j <= 9; j++)
			{
    
				a[k] = j;
				v = 0;
				for (i = 1; i <= k; i++)
					c[i] = 0;
				for (i = 1; i <= k; i++)
				{
    
					for (z = 0, t = 1; t <= k; t++)
					{
    
						u = a[i] * a[t] + z;
						z = u / 10;
						b[i + t - 1] = u % 10; //计算中间结果存于b数组
					}
						for (w = 0, m = i; m <= k; m++)
						{
    
							u = c[m] + b[m] + w; // 计算平方存于c数组 
							w = u / 10;
							c[m] = u % 10;
						}
					}
					for (i = 1; i <= k; i++)
						if (a[i] != c[i])
							v = 1;
					if (v == 0) break;
				}
			}
			if (v == 0 && a[n] != 0)
			{
    
				printf(" %d 结尾的%d位守形数： ", a[1], n);
				for (k = n; k >= 1; k--)
					printf("%d", a[k]);
				printf("\n");
			}
		}


	return 0;
}

结果：

//n位逐位整除数探索
int main()
{
    
	int i, j, n, r, t, s, a[100];
	printf("逐位整除数n位，请确定n：");
	scanf("%d", &n); 
	printf(" 所求%d 位逐位整除数：\n",n); 
	for (j = 1;j <= 100; j++)
	a[j] = 0;

	t = 0; 
	s = 0;
	i = 1;
	a[1] = 1;
	while (a[1] <= 9)
	{
    
		if (t == 0 && i < n)
			i++;
		for (r = 0, j = 1; j <= i; j++)
		{
    
			r = r * 10 + a[j];
			r = r % i;
		}
		//检测 i时是否整除 i 
		if (r != 0)
		{
    
			a[i] = a[i] + 1;
			t = 1;
			while (a[i] > 9 && i > 1)
			{
    
				a[i] = 0;
				i--;
				a[i] = a[i] + 1;
			}
		}
		else 
			t = 0;
		if (t == 0 && i == n)
		{
    
			s++;
			printf(" %d: ", s);
			for (j = 1; j <= n; j++)
				printf("%d", a[j]);
			printf("\n");
			a[i] = a[i] + 1;
		}
	}

			if (s == 0)
				printf("没有找到！\n");
			else
				printf(" 共以上%d个解。\n", s) ;

	return 0;
}

结果;

//递推探索n位逐位整除数
int main()
{
    
	int d, g, i, j,k, m, n, r, a[3000][30], b[3000][30];
	printf("请输入送位整除数的位数n：");
	scanf("%d", &n);
	g = 9; 
	for (j = 1; j <= g; j++)
		a[j][1] = j;
	for (k = 2; k <= n; k++)
	{
    

		m = 0;
		for(i = 1; i <= g; i++)
			for (j = 0; j <= 9; j++)
			{
    
				a[i][k] = j;
					for (r = 0, d =1; d <=k; d++ )
					{
    
						r = r * 10 + a[i][d];
						r = r % k;
					}
				if (r == 0)
				{
    
					m++;
					for (d = 1; d <= k; d++)
						b[m][d] = a[i][d];
				}

			}
		g = m;
		for (i = 1; i <= g; i++)
			for (d = 1; d <= k; d++)
				a[i][d] = b[i][d];
	}
	if (g > 0)
	{
    
		printf(" %d位逐位整数除数共%4d个", n,g);

		for (i =1; i <= g; i++)
		{
    
			printf(" %d: ", i);
			for (d = 1; d <= n; d++)
				printf("%d", a[i][d]);
			printf("\n");
		}
	}
	else
	{
    
		printf(" 无解！\n");
		return 0;
	}
	//return 0;
}

结果：