HDU5620——斐波那契数列的应用

题目描述：

KK's Steel

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 389 Accepted Submission(s): 184



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

Hint1+2+3=6 but 1+2=3  They are all different and cannot make a triangle.

题目链接

题意：

对于一条长为N(1≤N≤1018)米的钢管，最多可以锯成几根小钢管，使得锯成的钢管互不相等且均不能围成三角形。

解析：

一开始拿到题目的时候，只有短短的一句话，我们可能无从下手，当自己无从下手的时候，那么我们便可以枚举出前几种比较简单的情况，从而找出一般规律。由题意我们可以知道，锯成的钢管的长度至少为1，而任意的三根钢管又都不能围成三角形，并且锯成的钢管数目要最多，那么第二段钢管长度只能是2，同理，第三段钢管的长度只能是3，第四段是5，第五段是8.因此我们可以得出以下的数列：

1,2,3,5,8......

看到以上的数列，便立马反应过来了，这便是斐波那契数列。我们可以预处理出斐波那契数列的前n项和，然后再遍历即可。所以得出以下完整代码：

#include<cstdio>
#include<algorithm>
using namespace std;
typedef unsigned long long ull;
const int maxn = 100;
ull ans[maxn+5];
int main()
{
int T;
ull a1 = 1,a2 = 2;
ull  sum;
ans[1] = 1;
ans[2] = 3;
for(int i = 3; i < 100; i++)     //预处理出斐波那契数列前n项和
{
sum = a1 + a2;
ans[i] = ans[i-1] + sum;
a1 = a2;
a2 = sum;
}
scanf("%d",&T);
while(T--)
{
ull N;
scanf("%I64u",&N);
for(int i = 1; i < 100; i++)
{
if(N < ans[i])
{
printf("%d\n",i-1);
break;
}
}

}

return 0;
}

总结：当拿到一道没有头绪的题目时，可以从简单到复杂寻找规律，枚举当n=1,2,3,4...的情况，然后再总结出一般规律