UVA-100

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). 



Sample Input 


1  10 


100  200 


201  210 


900  1000 


Sample Output 


1  10  20 


100  200  125 


201  210  89 


900  1000  174 

题意：给出区间数， i 和 j， 找出其中的最大循环节长度（给定一个数，如果他是奇数就乘3加1， 如果他是偶数就除以2，一直到该数成为1为止）。

题解：该题的数据有点水， 暴力可以过，但是要注意给定的i 和 j ， i 和 j 的大小不确定， 最后输出的时候也要注意按照输入的顺序输出。

但是暴力最极端的例子， 是从1到最大， 这时候暴力实际上是超时的， 看了别人的思路， 自己原来不知道怎么打表， 就像备忘录（记忆化搜索）一样。打出表。

代码：

#include <iostream>#include<algorithm>#include<stdlib.h>#include<cstdio>#include<string.h>using namespace std;int number[1000000];int caculator(long long t) {    if(t == 1)        return 1;    if(t & 1) {        t += (t << 1) + 1;    }    else        t >>= 1;    if(t < 1000000) {        if(!number[t])            number[t] = caculator(t);        return  1 + number[t];    }    return  1 + caculator(t);}int main() {    int a, b, t1, t2, t3, Max;    memset(number, 0, sizeof(number));    for(int i = 1; i < 1000000; i++) {        number[i] = caculator(i);    }    while(scanf("%d%d", &a, &b) != EOF) {        Max = 0;        t1 = a + b;        t2 = a; t3 = b;        a = min(a, b);        b = t1 - a;        for(int i = a; i <= b; i++)            if(number[i] > Max)                Max = number[i];        printf("%d %d %d\n", t2, t3, Max);    }}