Interesting Integers Gym

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 first 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 first 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

这道题赛后看大佬们用的都是扩展欧几里得，懵。。。

我写的介个算法，坑点比较多，数据上细节比较多，稍不留神就容易错

奇葩思路A的，代码：

#include <stdio.h>
#include <string.h>

long long INF = 1e9+5;

//有数据用到了long long，就干脆把所有整数都开成longlong了，丢...
int main()
{
    long long t, x, a, b, re1, re2;
    long long data1, data2, t1, t2;
    long long aa[55], i, j;
    aa[1] = 1;
    aa[2] = 1;
    for(i=3; i<50; i++)
    {
        aa[i] = aa[i-1] + aa[i-2];
    }
    scanf("%lld", &t);

    while(t--)
    {
        scanf("%lld", &x);
        for(i=48; i>=0; i--)
        {
            if(aa[i]+aa[i+1]<=x)   //这句剪枝不能没有，没有会T，
                 //想一下，这样就可以减少好多没有意义的80000次循环，何乐而不为？
            {
                for(j=1; j<=80000; j++)  //在不T的条件下尽量扩大范围
                {
                    if((x-aa[i+1]*j)%aa[i]==0&&(x-aa[i+1]*j)/aa[i]>0&&(x-aa[i+1]*j)/aa[i]<=j) break;
                }
                if(j<=80000) break;
            }
        }
        printf("%lld %lld
", (x-aa[i+1]*j)/aa[i], j);
    }
    return 0;
}