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
参考代码
#include <stdio.h>
int F(int i);
int main()
{
int a,b,i,temp=0,temp1,temp2,a1,a2;
while(scanf("%d %d",&a,&b)==2)
{
a1=a;
a2=b;
if(a>b)
{
temp2=a;
a=b;
b=temp2;
}
for (i=a;i<=b;i++)
{
temp1=F(i);
//printf("%dn",temp1);
if(temp<temp1)
temp=temp1;
}
printf("%d %d %dn",a1,a2,temp);
temp=0;
}
}
int F(int i)
{
int temp=i,j=1;
while(1)
{
if(temp==1)
return j;
if(temp%2==0)
{
temp=temp/2;
}
else
{
temp=temp*3+1;
}
j++;
}
}
解析
暂无
