HDU 5620 KK's Steel（思维题）

KK’s Steel

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 737 Accepted Submission(s): 345

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：给出一个钢管，你可以把它拆分为任意个数，并且要满足一下要求。
应该满足的条件为：
1.这组数的任意两个数都不相等。
2.并且任意三个不能形成三角形。
问你可以分为多少个数
求出划分的个数。
首先从6开始推起，6可以划分为1，2，3。
7的话可以划分为1，2，4。
8的话可以分为1，2，5。
9的话可以分为1，2，6。
10的话可以分为2，3，5或者1，4，5。
11的话可以分为1，2，3，5。4个
相信到现在大家应该可以发现规律了，如果还没有发现的话，可以先看下代码。
下面是AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

long long a[200];
long long b[200];
void init()
{
    a[1]=1;
    a[2]=2;
    for(int i=3;i<=200;i++)
    {
        a[i]=a[i-1]+a[i-2];
    }
    long long sum=0;
    for(int i=1;i<=200;i++)
    {
        sum+=a[i];
        b[i]=sum;
    }
}
int main()
{
    int t;
    init();
    scanf("%d",&t);
    while(t--)
    {
        long long n;
        scanf("%lld",&n);
        int flag=0;
        for(int i=1;i<190;i++)
        {
            if(b[i]==n)
            {
                flag=i;
                break;
            }
            else if(b[i]>n)
            {
                flag=i-1;
                break;
            }
        }
        if(flag<3)
        {
            flag=0;
        }
        printf("%d\n",flag);
    }
    return 0;
}