hdoj-5620-KK's Steel

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. 
    
    
    
    
     
   
   
   
   

长为n的钢....拆分，要求每两段不一样，没三段不能构成三角，

肯定从1开始嘛，然后是2，然后不能构成三角形所以1+2=3，然后你会发现是斐波拉契数列

嗯...

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 1000 + 10;
const int maxt = 75 + 10;
typedef long long ll;
ll a[maxt];
int main()
{
    a[1] = 1;
    a[2] = 2;
    for (int i = 3 ; i < maxt; ++i)a[i] = a[i-1] + a[i-2];
    int n;
    cin >> n;
    while(n--){
        ll x;
        cin >> x;
        int i;
        ll sum = 0;
        for (i = 1; i < maxt; ++i){
            sum += a[i];
            if (sum > x)break;
        }
        if (x == 2 || x == 1){cout << 1 << endl;continue;}
        cout << i-1<< endl;
    }
    return 0;
}