The 3n + 1 problem

code

4 年前

The 3n + 1 problem

时间： 1ms 内存：64M

描述：

Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example, the following sequence of numbers will be generated for n = 22:
22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for every integer n. Still, the conjecture holds for all integers up to at least 1, 000, 000.

For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.

输入：

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0.

输出：

For each pair of input integers i and j, output i, j in the same order in which they appeared in the input and then the maximum cycle length for integers between and including i and j. These three numbers should be separated by one space, with all three numbers on one line and with one line of output for each line of input.

示例输入：

示例输出：

1 10 20
100 200 125
201 210 89
900 1000 174

提示：

参考答案（内存最优[748]）：

#include <stdio.h>
int F(int i);
int main()
{
	int a,b,i,temp=0,temp1,temp2,a1,a2;
	while(scanf("%d %d",&a,&b)==2)
	{
                  a1=a;
                  a2=b;
                  if(a>b)
                  {
                      temp2=a;
                      a=b;
                      b=temp2;
                  }
		for(i=a;i<=b;i++)
		{
			temp1=F(i);
			//printf("%d\n",temp1);
			if(temp<temp1)
				temp=temp1;
		}
		printf("%d %d %d\n",a1,a2,temp);
		temp=0;
	}
}

int F(int i)
{
	int temp=i,j=1;
	while(1)
	{
		if(temp==1)
			return j;
		if(temp%2==0)
		{
			temp=temp/2;
		}
		else
		{
			temp=temp*3+1;
		}
		j++;
	}
}

参考答案（时间最优[0]）：

#include"stdio.h"
int a[105][105];
char s[105][105];
int main()
{
	int m,n,i,j,c=0;
	while(scanf("%d %d",&n,&m)==2&&(n||m))
	{
		getchar();
		for(i=0;i<=n;i++)
			for(j=0;j<=n;j++)
				a[i][j]=0;
		c++;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				scanf("%c",&s[i][j]);
			if(s[i][j]=='*')
			{
				a[i][j]=-1;
				if(a[i-1][j-1]!=-1)a[i-1][j-1]++;
				if(a[i][j-1]!=-1)a[i][j-1]++;
				if(a[i-1][j]!=-1)a[i-1][j]++;
				if(a[i+1][j+1]!=-1)a[i+1][j+1]++;
				if(a[i-1][j+1]!=-1)a[i-1][j+1]++;
				if(a[i+1][j-1]!=-1)a[i+1][j-1]++;
				if(a[i][j+1]!=-1)a[i][j+1]++;
				if(a[i+1][j]!=-1)a[i+1][j]++;
			}
			}
			getchar();
		}
		printf("Field #%d:\n",c);
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				if(a[i][j]==-1)printf("*");
				else printf("%d",a[i][j]);
			}
			printf("\n");
		}
		printf("\n");
	}
 	return 0;
}

题目和答案均来自于互联网，仅供参考，如有问题请联系管理员修改或删除。