hdu 5620 KK's Steel（推理）

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)

 

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 ！

思路：

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！

 

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 using namespace std;
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 1000000
 #define inf 1e12
 ll n;
 ll f[N];
 void init(){
     f[]=;
     f[]=;
     for(ll i=;i<N;i++){
         f[i]=f[i-]+f[i-];
     }
 }
 int main()
 {
     init();
     int t;
     scanf("%d",&t);
     while(t--){
         scanf("%I64d",&n);
         ll sum=;
         ll i;
         for(i=;i<N;i++){
             sum+=f[i];
             if(sum>n){
                 break;
             }
         }
         if(n== || n==){
             printf("1\n");
             continue;
         }
         printf("%I64d\n",i-);
     }
     return ;
 }