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

16

 

Sample Output

3
Hint
1+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