一、題目信息
The 3n + 1 problem
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 44458 Accepted: 13979
Description www.2cto.com
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 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 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.
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
二、算法分析
首先，開一個10000的數組，將所有1~9999的cycle length求出來，之後再根據輸入的范圍求最大值。這裡要特別注意的是，輸入的范圍，並不一定是i<j的，還有可能是i>j，此時，應該要將i,j交換之後再求最大值。
我們在求cycle length的時候，可以利用我們已經求過的length。
三、參考代碼
[cpp]
#include <stdio.h>
#include <string.h>
int seq[10000],len;
int count_q(int n)
{
int res = 0;
while(1)
{
if(n == 1)
{res++;break;}
else if(n < 10000 && seq[n])
{res += seq[n];break;}
else if(n&1)
{res++;n = 3*n + 1;}
else
{res++;n = n>>1;}
}
return res;
}
int maxww(int beg,int end)
{
int tmp,i;
int res = seq[beg];
if(beg > end)
tmp = beg,beg = end,end = tmp;
for(i = beg+1; i <= end; i++)
seq[i] > res ? res = seq[i] : 0;
return res;
}
int main(int argc , char * argv[]){
int i,j,beg,end;
memset(seq,0,10000*sizeof(int));
for(i = 1;i < 10000 ; i++){
if(!seq[i]){
seq[i] = count_q(i);
for(j = i << 1,len = seq[i]+1 ; j < 10000; j = j << 1,len++)
!seq[j] ? seq[j] = len :0 ;
}
}
while(scanf("%d%d",&beg,&end)!=EOF)
printf("%d %d %d\n",beg,end,maxww(beg,end));
return 0;
}
