ACM练习3*n+1问题

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 10100 200201 210900 1000

样例输出

1 10 20100 200 125201 210 89900 1000 174

提示

杭电1032

这是题目，大概意思是对于一个数，如果这个数的是偶数，就除以2，如果是奇数就乘以3然后加上1，直到最终结果为1为止，这个规律对于任何数都成立，记录从开始到1的步骤，要求输入一个数，i和j，求出这两个数之间步骤最少的一个。没有告诉一共输入多少组。

我应该注意的是：不是把所有的输入都输入完成，然后做个总的输出，而是每个输入对应一组输出。

#include<stdio.h>int main(){       int i,j,k,k2,max=0,count=0,tem,j1,j2;       while(scanf("%d %d",&i,&j)!=EOF)//这里不能改成while（1），改后会提示超出限制       {           max=0;           j1=i;           j2=j;           if(i>j)//这里必须判断大小，不能看例子都是i小于j的所以就自以为是i<j           {                tem=i;                i=j;                j=tem;           }           for(k=i;k<=j;k++)           {               for(k2=k,count=1;count==0||k2!=1;count++)               {                   if(k2%2==0)                   {                       k2=k2/2;                   }                   else                   {                       k2=k2*3+1;                   }                }                 if(count>max)                {                    max=count;                }           }            printf("%d %d %d\n",j1,j2,max);       }    return 0;}