斐波那契首项 BAPC2014 I&&HUNNU11589:Interesting Integers

题目大意：给定一个数n，按斐波那契数规则生成，求最小的b，并使a最小。G1 = a , G2 = b; b >= a.

思路分析：

因为n<10^9,所以斐波那契数列的项数不可能超过46.题目转换为求ax+by=n的最小的x，y。根据欧几里得

#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;

#define maxn 99999999
#define mem(a) memset(a , 0 , sizeof(a))
#define LL __int64
LL fib[65];

void init()
{
    fib[0] = 1;
    fib[1] = 1;
    fib[2] = 2;
    for(int i = 3 ; i < 65 ; i ++)
    {
        fib[i] = fib[i-1] + fib[i-2];
    }
    return ;
}

void gcd2(LL a , LL b , int &d ,LL &x , LL &y)
{
    if(!b)
    {
        d = a;
        x = 1 ;
        y = 0;
    }
    else
    {
        gcd2(b , a%b , d , y , x);
        y -= x*(a/b);
    }
}


int main()
{
    int t;
    LL n;
    init();
    scanf("%d" , &t);
    while(t-- )
    {
        scanf("%I64d" , &n);
        int pos;
        pos = 65;
        LL ansx = (LL)maxn*1000, ansy = (LL)maxn*1000;
        LL x , y ;int d;
        while(pos > 2)
        {
            LL a = fib[pos - 2], b = fib[pos - 1];
            pos -- ;
            gcd2(a , b , d , x , y);
            x *= n/d ;
            y *= n/d;
            LL t1 = b / d;
            LL t2 = a / d;
            x = (x % t1 + t1) % t1;     //找到最小正整数x
            y = (n - a * x) / b;     //对应的y；
            
            ////////////并没有看懂这个
             ll   k = (y-x-1) / (ma+mb);//优化  
            x += ma*k, y -= mb*k;  
           /* int abc;
            if(x <= 0)
            {
                abc = abs(x / b);
                if(x % b != 0)  abc ++;
                x = x + abc * b / d;
                y = y - abc * a / d;
            }
            if(y <= 0)
            {
                abc = abs(y / a);
                if(y % a != 0)  abc ++;
                x = x - abc * b / d;
                y = y + abc * a / d;
            }
            if(x < 0 || y + x < 0) continue;
            if( x == 0) swap(x , y);
            else if(y != 0 && x > y) swap(x , y);
            if(x == 0 && y == 0) continue;*/
            while(y >= x && y < ansy)
            {
                if(y < ansy || (y == ansy && x < ansx) && x > 0) ansx = x , ansy = y;
                x += t1  , y -= t2;
            }
        }
        if(ansx == 0) ansx = ansy ;
        printf("%I64d %I64d\n" , ansx ,ansy);
    }
}