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;
}