1409061046-hd-The 3n + 1 problem

The 3n + 1 problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1044    Accepted Submission(s): 483

 

Problem Description

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

Consider the following algorithm:


    1.      input n

    2.      print n

    3.      if n = 1 then STOP

    4.           if n is odd then   n <- 3n + 1

    5.           else   n <- n / 2

    6.      GOTO 2


Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)

Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.

For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.

Input

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.

You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.

You can assume that no opperation overflows a 32-bit integer.

Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).

Sample Input

1 10100 200201 210900 1000

Sample Output

1 10 20100 200 125201 210 89900 1000 174

题目大意

       给出两个数，根据题目给出的变化求这个区间内步骤最多的一个数的变化步骤。

解题思路

       要注意一点，题目给出的两个数并没有注明谁大谁小，所以一定要比较一下。

代码

<span style="font-size:18px;">#include<stdio.h>int s[1100000];int main(){int n,m;int a,b;int i,j,k,max;while(scanf("%d%d",&n,&m)!=EOF){if(n>m){a=m;b=n;}else{a=n;b=m;}//给出的两个数并没有说谁大谁小，所以需要比较。     for(i=a;i<=b;i++)    {    s[i]=1;    j=i;    while(j!=1)    {    if(j%2==1)       j=j*3+1;     else       j/=2;    s[i]++;    }    }max=-1;for(i=a;i<=b;i++)    if(s[i]>max)        max=s[i];printf("%d %d %d\n",n,m,max);}return 0;}</span>