Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 56636 | Accepted: 18061 |
Description
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
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
Sample Input
1 10 100 200 201 210 900 1000
Sample Output
1 10 20 100 200 125 201 210 89 900 1000 174
有個坑點就是i不一定小於j,所以要記得比較大小,最後輸出的時候要原樣輸出
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int n,m;
int main()
{
while(~scanf("%d%d",&n,&m))
{
int Max = -1;
int temp_n = n;
int temp_m = m;
if(n > m)
{
int temp = n;
n = m;
m = temp;
}
for(int i=n;i<=m;i++)
{
int ans = 1;
int k = i;
while(k != 1)
{
ans++;
if(k % 2 == 1)
{
k = 3 * k + 1;
}
else
{
k /= 2;
}
}
Max = max(Max,ans);
}
printf("%d %d %d\n",temp_n,temp_m,Max);
}
}