100 - The 3n+1 problem

Time limit 3.000 seconds

Problems in Computer Science are often classified as belonging to a certain class of problrms(e.g.,NP,Unsolvable,Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not konwn 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 sunch 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 maximum cycle length over all numbers between and including the 1 is printed .For a given n this is called the cyccle-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 onsist of a srries 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 maxmum cycle length over all integers between and including i and j.

You can assume that no operation overflows a 32-bit integer.

Ouput

For each pair of input integers i and j you should output i,j,and the maxmum cycle length for integers between and including i and j.There 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 inout,The integers i ang j must appear in the output int the same order in which they appeared in the input and be followed by the maxmum 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

代码1：此代码时间超限，但是打表策略体现在其中。

#include<stdio.h>
#include<iostream>
using namespace std; 
int a[1000001];

int num(int n)
{
	int i=1;//数值自身计入1 
	while(n!=1)
	{
		if(n&1)//奇数 
		n=3*n+1;
		else
		n>>=1;//将n对应的二进制右移一位，对应十进制减半； 
		i++; 
	}
	return i;
 } 
 
 void numlen()//存储每个数得计算次数 
 {
 	int i;
 	for(i=1;i<1000001;i++)
 	a[i]=num(i); 
 }
 
 int main()
 {
 	numlen(); 
 	int x,y,n,maxnum,min,max,i;
 	while(scanf("%d%d",&x,&y)!=EOF)
 	{
 		maxnum=0;
 		min=x<y?x:y;
 		max=x>y?x:y;
 		for(i=min;i<max;i++)
 		{
 			n=a[i];
 			if(n>maxnum)
 			maxnum=n;
		 }
		 printf("%d %d %d\n",x,y,maxnum);
	 }
	 return 0;
 }

代码2：在UVaOJ上运行通过的代码：

#include<cstdio>
#include<algorithm>
typedef long long LL;
const int maxn = 1000001;
int data[maxn];//打表 
using namespace std;
#include<map>

int main()
{
	for(int d=1;d<maxn;d++)
	{
		int cnt=1;
		LL i=d;
		while(i>1)
		{
			if(i&1) i=3*i+1;
			else i>>=1;
			cnt=cnt+1;
		}
		data[d]=cnt;
	}//实现打表 
	int i,j;
	while(scanf("%d%d",&i,&j)!=EOF)
	{
		int m=i,n=j,min,max;
		min = i<j?i:j;
		max = i>j?i:j;
		int maxn =0;
		for(int d=min;d<=max;d++)
		
			if(data[d]>maxn) maxn=data[d];
			printf("%d %d %d\n",m,n,maxn);
		
	}
}

什么是打表：

打表是一种典型的空间换时间的技巧，一般指的是：将所有可能用到的结果事先计算出来，这样后面需要用到时可以直接查表获得。