YTU 1098: The 3n + 1 problem

1098: The 3n + 1 problem

时间限制: 1 Sec  内存限制: 64 MB
提交: 368  解决: 148

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

总是望着曾经的空间发呆，那些说好不分开的朋友不在了，转身，陌路。 熟悉的，安静了， 安静的，离开了， 离开的，陌生了， 陌生的，消失了， 消失的，陌路了。

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int a,b,i,j=0,m=0,c=0;
    for(; ~scanf("%d%d",&a,&b); m=0)
    {
        for(c=a>b?b:a; c<=(a>b?a:b); c++)
        {
            i=c,j=0;
            for(; i!=1; j++)
                if(i%2==0)i/=2;
                else i=i*3+1;
            m=j>m?j:m;
        }
        printf("%d %d %d
",a,b,m+1);
    }
    return 0;
}