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一个问题

The Problem

Consider the following algorithm:

1. input n 2. print n 3. if n = 1 then STOP 4. if n is odd then 5. else 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 (including the 1). 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.

The 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 1,000,000 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 opperation overflows a 32-bit integer.

The 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

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174

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第 2 楼

得分:0

这是我的代码，不能过。不知道是什么原因，请大家帮忙找找
#include <iostream.h>
#include <fstream.h>

const int maxn=1000000;

unsigned long table[maxn]={0};

void count()
{
table[1]=1;
for(long i=2;i<maxn;i++){
long k=i, s=0;
while(1){
if(k%2)
k=k*3+1;
else
k=k/2;
s++;
if(table[k]){
table[i]=s+table[k];
break;
}
}
}
}

unsigned long compare(long a,long b)
{
long max=0;
for(long i=a; i<=b; i++)
if(table[i]>max)max=table[i];
return max;
}

int main()
{
long i, j;
count();
while(cin){
cin>>i>>j;
cout<<i<<' '<<j<<' '<<compare(i,j)<<endl;
}
return 0;
}

第 3 楼

得分:0

你定义的mixn=1000000，数据溢出．
建议你的输入输出使用提示，这样你会分清楚．
修改了一下，没有问题的：
#include <iostream.h>
#include <fstream.h>

const long int maxn=1000;

unsigned long table[maxn]={0};

void count()
{
table[1]=1;
for(long i=2;i<maxn;i++)
{
long int k=i;
long int s=0;
while(1)
{
if(k%2)
k=k*3+1;
else
k=k/2;
s++;
if(table[k])
{
table[i]=s+table[k];
break;
}
}
}
}

unsigned long compare(long a,long b)
{
long max=0;
for(long i=a; i<=b; i++)
if(table[i]>max)max=table[i];
return max;
}

int main()
{
long i, j;
count();
while(cin){
cout<<"请输入数据"<<endl;
cin>>i>>j;
cout<<"输出为:"<<endl;
cout<<i<<' '<<j<<' '<<compare(i,j)<<endl;
}
return 0;
}
输出的内容有：
请输入数据
1 10
输出为:
1 10 20
请输入数据
100 200
输出为:
100 200 125
请输入数据
201 210
输出为:
201 210 89
请输入数据
900 1000
输出为:
900 1000 174
请输入数据