HDU 5620 KK's Steel

KK's Steel

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 20    Accepted Submission(s): 13

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)

 

大体题意：

给你一个长度为N的钢管，问最多切成几个钢管，使得这些钢管不能围成三角形，并且不能有相同长度 ！

分析：

开始想感觉好麻烦，其实仔细想想吧：

要想使得钢管尽量多，那肯定从1开始吧，有了1，且不能重复，那肯定找2吧，而且三个钢管不能围成，三角形，那直接找a1 + a2 = a3的情况不就恰好不能围成三角形吗。所以很明显，这是一个a1 = 1，a2 = 2的斐波那契数列，找到第一个i  是的前i项和大于N，即可！特殊判断N = 1，N=2即可！他们都是1！感觉这里题目阐述有点不理解！！

#include<iostream>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<string>
#include<sstream>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<cstdlib>
#define mem(x) memset(x,0,sizeof(x));
#define mem1(x) memset(x,-1,sizeof(x));
using namespace std;
const int maxn = 1000 + 10;
const int maxt = 75 + 10;
const double eps = 1e-8;
const int INF = 1e8;
const double pi = acos(-1.0);
typedef long long ll;
typedef unsigned long long llu;
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;
}