BestCoder Round #71 (div.2) （hdu 5620 菲波那切数列变形）

Posted 2020-06-13

KK‘s Steel

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 350    Accepted Submission(s): 166

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.

 

要求：给一个长整形数n让你把n分成若干份，1、任意两份长度不能相等 2、任意三分不能组成三角形

 

分析：因为斐波那契数列中的数满足此要求，所以从这个方面入手，只要分成的数满足a+b<=c即可，所以我们在数列1  2  3  5  8  13  21 ......的数中选取，如果前i项的和等于n输出i如果前i项的和大于n输出i-1

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define MAX 100100
#define INF 0x3f3f3f
#define LL long long
using namespace std;
LL fb[10010];
LL f[1001];
void biao()
{
	LL i,j;
	fb[1]=1;
	fb[2]=2;
	for(i=3;i<120;i++)
	    fb[i]=fb[i-1]+fb[i-2];
	f[1]=fb[1];
	for(i=2;i<120;i++)
	    f[i]=f[i-1]+fb[i];
}
int main()
{
	int t,i,j;
	LL n;
	biao();
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		for(i=1;i<120;i++)
		{
			if(n==f[i])
			{
				printf("%d\n",i);
				break;
			}
			if(n<f[i])
			{
				printf("%d\n",i-1);
				break;
			}
		}
	}
	return 0;
}