题号：UVA100
时间限制：C/C++ 3秒，其他语言6秒
空间限制：C/C++ 0K，其他语言0K
64bit IO Format: %lld
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 before and including
the 1 is printed. 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 and including both 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 10,000 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 operation 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).
大概意思：‎请考虑以下算法以生成数字序列。以整数 n 开头。如果 n 为偶数，则除以 2。如果 n 为奇数，则乘以 3 并加 1。使用新值 n 重复此过程，并在 n = 1 时终止。例如，将为 n = 22 生成以下数字序列： 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 推测（但尚未证明）对于每个整数 n，此算法将终止于 n = 1。尽管如此，这个猜想仍然适用于至少1， 000， 000的所有整数。对于输入 n，n 的循环长度是生成的数量，最多包括 1。在上面的示例中，循环长度 22 为 16。给定任意两个数字 i 和 j，您将确定 i 和 j 之间所有数字（包括两个端点）的最大循环长度。‎
Sample Input
1 10
100 200
201 210
900 1000
‎输入将由一系列整数对 i 和 j 组成，每行一对整数。所有整数将小于 1，000，000 且大于 0。‎
Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174
#include<iostream>
using namespace std;

int main(){
    long long i,j,k,s,num,max,t;
    while((scanf("%lld %lld",&i,&j))!=EOF){
        printf("%lld %lld ",i,j);
        if(i>j){
            t = i;
            i = j;
            j = t;
        }
        max = 1;
        for(i = k; j<=j; k++){
            num = k;
            s = k;
            while(s!=1){
                if(s%2){
                    s=3*s+1;
                    num++;
                }else{
                    s = s/2;
                    num++;
                }
            }
            if(num>max) max = num;
        }
        printf("%lld\n",max); 
    }
    return 0;
}