Undoubtedly you know of the Fibonacci numbers. Starting with
F1 = 1 and F2 = 1, every next number is the sum of the two
previous ones. This results in the sequence 1, 1, 2, 3, 5, 8, 13, . . ..
Now let us consider more generally sequences that obey the
same recursion relation
Gi = Gi?1 + Gi?2 for i > 2
but start with two numbers G1 ≤ G2 of our own choice. We shall
call these Gabonacci sequences. For example, if one uses G1 = 1
and G2 = 3, one gets what are known as the Lucas numbers:
1, 3, 4, 7, 11, 18, 29, . . .. These numbers are – apart from 1 and 3 –
different from the Fibonacci numbers.
By choosing the ?rst two numbers appropriately, you can get
any number you like to appear in the Gabonacci sequence. For
example, the number n appears in the sequence that starts with 1
and n ? 1, but that is a bit lame. It would be more fun to start with numbers that are as small
as possible, would you not agree?
Input
On the ?rst line one positive number: the number of test cases, at most 100. After that per test
case:
? one line with a single integer n (2 ≤ n ≤ 109
): the number to appear in the sequence.
Output
Per test case:
? one line with two integers a and b (0 < a ≤ b), such that, for G1 = a and G2 = b,
Gk = n for some k. These numbers should be the smallest possible, i.e., there should be
no numbers a
0 and b
0 with the same property, for which b
0 < b, or for which b
0 = b and
a
0 < a.
Sample in- and output
Input
5
89
123
1000
1573655
842831057
Output
1 1
1 3
2 10
985 1971
2 7
解题：斐波那契第n项：a[n]=f[n-1]*x+f[n]*y; // f[n]：f[1]=0，f[2]＝1；的斐波那契数列。枚举n与y看能否整除f[n-1]，且除数<=y。x:斐波那契第一项，y:斐波那契第二项。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define Max(a,b) (a>b?a:b)
using namespace std;
#define ll long long
int main (void)
{
int f[1005] , ans ;
int y ,x;
f[1]=0;
f[2]=1;
int i=3;
for( i=3; i<=46; i++)
{
f[i]=f[i-1]+f[i-2];
}
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&ans);
if(ans==1||ans==2)
{
printf("1 1\n");
continue;
}
bool bb=0;
for(int i=45 ; i>2&&!bb; i--)
for(int ty=1; ty<=1000000; ty++)
if(ty*f[i]+f[i-1]>ans)
break;
else if((ans-ty*f[i])%f[i-1]==0&&(ans-ty*f[i])/f[i-1]<=ty)
{
y=ty , x=(ans-ty*f[i])/f[i-1] , bb=1;
break;
}
printf("%d %d\n",x,y);
}
return 0;
}
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