ACM题目1095: The 3n + 1 problem

题目描述

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.

样例输入

1 10
100 200
201 210
900 1000

样例输出

1 10 20
100 200 125
201 210 89
900 1000 174
思路：
有了初始数字i和循环次数j
就可以用if判断来得到最后的i值
我自己编写的程序显示答案错误
错误代码：

#include<stdio.h>
int main()
{
int i,j;//i用来保存起始的数字，j用来保存遍历的次数
scanf("%d %d",&i &j);
int m;//m用来保存更新后的i值
while(j--)
{
if(i%2==0)
{
i=i/2;
}
if(i%2!=0)
{
i=i*3+1;
}
}
m=i;
printf("%d %d %d",i j m);
return 0;

}

错误提示：

错误原因：

当i执行一定次数之后数字变为1程序即终止，但是给了i和j的值 不能确定遍历j次后程序是否已经终止

所以要对i和j的值进行大小的判断,忽略判断的话就会导致有一部分特殊情况没有考虑到

正确代码：

#include<stdio.h>

int dfs(int a)
{
   int i=1;
   while(a!=1)
   {
       if(a%2==0)
       {
           i++;
           a/=2;
       }
       else
       {
           i++;
           a=(a*3)+1;
       }
   }
   return i;

}
int  main()
{
    
    int i,j,t;//t作为临时变量用来判断i j大小
    int num;//num用来计算比较的次数
    int k;
    
    while(scanf("%d%d",&i,&j)!=EOF)
    {
        printf("%d %d ",i,j);
        if(i>j)
        {
           
            t=i;
            i=j;
            j=t;
        }
	    int max=0;//循环长度初始化长度
	    for(k=i;k<=j;k++)
	    {
	        num=dfs(k);
	        if(num>max)
	        {
	            max=num;
	        }
	        
	    }
	    printf("%d
",max);
	}
    return 0;
}

反思：

考虑问题要充分一些

另外敲代码的时候要仔细一些 把每一个字母都敲正确，防止出现一些因为粗心而出现的不该犯的错误