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
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！
1 #pragma comment(linker, "/STACK:1024000000,1024000000")
2 #include<iostream>
3 #include<cstdio>
4 #include<cstring>
5 #include<cmath>
6 #include<math.h>
7 #include<algorithm>
8 #include<queue>
9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 using namespace std;
15 #define ll long long
16 #define eps 1e-10
17 #define MOD 1000000007
18 #define N 1000000
19 #define inf 1e12
20 ll n;
21 ll f[N];
22 void init(){
23 f[1]=1;
24 f[2]=2;
25 for(ll i=3;i<N;i++){
26 f[i]=f[i-2]+f[i-1];
27 }
28 }
29 int main()
30 {
31 init();
32 int t;
33 scanf("%d",&t);
34 while(t--){
35 scanf("%I64d",&n);
36 ll sum=0;
37 ll i;
38 for(i=1;i<N;i++){
39 sum+=f[i];
40 if(sum>n){
41 break;
42 }
43 }
44 if(n==1 || n==2){
45 printf("1\n");
46 continue;
47 }
48 printf("%I64d\n",i-1);
49 }
50 return 0;
51 }
hdu 5620 KK's Steel（推理）