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[1]=1;
    f[2]=2;
    for(ll i=3;i<N;i++){
        f[i]=f[i-2]+f[i-1];
    }
}
int main()
{
    init();
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%I64d",&n);
        ll sum=0;
        ll i;
        for(i=1;i<N;i++){
            sum+=f[i];
            if(sum>n){
                break;
            }
        }
        if(n==1 || n==2){
            printf("1
");
            continue;
        }
        printf("%I64d
",i-1);
    }
    return 0;
}