hd 5620 KK's Steel（斐波那契数列）_our lovely kk has a difficult mathematical problem-优快云博客

本文介绍了一道数学问题：如何将长度为N的钢材切割成尽可能多的段，每段长度各不相同且任意三段无法构成三角形。文章给出了使用斐波那契数列解决问题的代码实现。

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KK's Steel

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1009 Accepted Submission(s): 487

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.
题意：给你一根长为n的木棒，怎你最多能把它分成多少段，并且要求任意两段长度不相等，任意三段不能组成三角形
思路：通过观察，只能分成 1  2  3  5  8 13  21 ..........         即斐波那契数列
 #include<cstdio>
#define LL long long int
int main()
{
    int t;
    LL n;
	LL f[100];
    f[1] = 1;
    f[2] = 2;
    for(int i = 3 ; i <= 85 ; i++)
    {
        f[i] = f[i-1] + f[i-2];
    }
    scanf("%d",&t);
    while(t--)
    {
        LL sum = 0;
        int falg = 0;-
        scanf("%I64d",&n);
        for(int i = 1 ; i <= 85 ; i++){
            sum += f[i];
            if(sum >= n)
            {
                falg = i;
                break;
            }
        }
        if(sum != n)
            falg--;
        printf("%d\n",falg);
    }
    return 0;
}