题目
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
注意！
i, j的大小顺序
代码
i,j=map(int,input().split())
print(i ,j ,end=' ')
if i<j:
pass
else:
i,j=j,i
while True:
try:
l=[]
for k in range(i,j+1):
c=1
while k!=1:
if k % 2 == 0:
k = k / 2
c += 1
else:
k = k * 3 + 1
c += 1
l.append(c)
print(max(l))
i,j=map(int,input().split())
print(i, j , end=' ')
if i < j:
pass
else:
i, j = j, i
except ValueError:
break
|