All in one 1086: Jiaogu conjecture

【 Title Description 】

It is called Jiaogu conjecture , For any positive integer , If it's odd , Then take 3 Add 1, If it's even , Divide by 2, The results are repeated according to the above rules , In the end, you can always get 1. Such as , Suppose the initial integer is 5, The calculation process is as follows 16、8、4、2、1. The program requires an integer , Will be processed to 1 The process output .

【 Input 】

A positive integer N(N <= 2,000,000).

【 Output 】

Enter an integer from to 1 Steps for , Each step is a line , The calculation process is described in each part . The last line outputs "End". If the input is 1, Direct output "End".

【 sample input 】

5

【 sample output 】

5*3+1=16
16/2=8
8/2=4
4/2=2
2/2=1
End

【 Tips 】

no

#include<stdio.h>
int main()
{
	int n;
	scanf("%d", &n);
	int temp = n;
	if (n == 1)
	{
		printf("End");
	}
	else
	{
		if (n % 2 != 0)
		{
			n = n * 3 + 1;
			printf("%d*3+1=%d\n", temp, n);
		}
		temp = n;
		while (1)
		{
			n /= 2;
			if (n != 0)
			{
				printf("%d/2=%d\n", temp, n);
				temp /= 2;
			}
			else
			{
				printf("End");
				break;
			}
			if (n % 2 != 0&&n!=1)
			{
				temp = n * 3 + 1;
				printf("%d*3+1=%d\n", n, temp);
				n = temp;
			}
			if (n == 1)
			{
				printf("End");
				break;
			}
		}
	}
	return 0;
}