【杭电oj】5620 - KK's Steel（递推）

KK's Steel

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 650    Accepted Submission(s): 301

Problem Description

Our lovely KK has a difficult mathematical problem:he has a  N(1≤N≤1018)  meters steel,he will cut it into steels as many as possible,and he doesn't want any two of them be the same length or any three of them can form a triangle.

 

Input

The first line of the input file contains an integer  T(1≤T≤10) , which indicates the number of test cases.

Each test case contains one line including a integer  N(1≤N≤1018) ,indicating the length of the steel.

 

Output

For each test case, output one line, an integer represent the maxiumum number of steels he can cut it into.

 

Sample Input

   
   
   
   

    
    
    
    1
6
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3


    
    
    
    
 
     
 
      Hint 
     
1+2+3=6 but 1+2=3 They are all different and cannot make a triangle. 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #71 (div.2)

类似斐波那契数列，递推过去就行了，时间上足够，注意数据类型就行了。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
long long f[1000000];
int main()
{
	f[1] = 1;
	f[2] = 2;
	for (int i = 3 ; i <= 1000 ; i++)
		f[i] = f[i-1] + f[i-2];
	int u;
	long long n;
	long long sum;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%lld",&n);
		sum = 0;
		for (int i = 1 ;; i++)
		{
			sum += f[i];
			if (sum == n)
			{
				printf ("%d\n",i);
				break;
			}
			else if (sum > n)
			{
				printf ("%d\n",--i);
				break;
			}
		}
	}
	return 0;
}