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
#include <iostream>
using namespace std;
//求循环长度
int maxlength(int n){int t = 0;while(n != 1){if(n%2 == 0) n = n/2; else n = 3*n+1; t++;}return t+1;
}
int main()
{int m,n,i,t;while(cin>>m>>n){cout<<m<<' '<<n<<' ';
//当m大于n，需要对调。而输出的还得是原来的顺序。 if(m>n){t = m; m = n; n = t;}t = 0;for(i = m;i<=n;i++){if(t<maxlength(i)) t = maxlength(i);} cout<<t<<endl; }return 0;
}