BAPC2014 I&&HUNNU11589:Interesting Integers

题意：

对于我们已知的斐波那契数列，现在给出一个n，要我们求出一个新的斐波那契数列起始项，使得n能在新的斐波那契数列中，要求起始项y最小

思路：

我们知道

a3 = a1+a2

a4 = a1+2*a2

a5 = 2*a1+3*a2

a6 = 3*a1+5*a2

可以得到

an = fib[n-2]*a[1]+fib[n-1]*a[2]

然后我们只需要枚举就可以了，猜测a[2]不会超过10W，然后居然也过了

队友用数学方法解的，那个才是正解，暴力是邪门歪道，囧

http://blog.csdn.net/u010579068/article/details/47452185

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define ls 2*i
#define rs 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define ULL unsigned long long
#define N 100005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
const int mod = 1000000007;

LL fib[50]= {0,0,1};
LL n ;

int main()
{
    LL i,j,k;
    for(i = 3; i<50; i++)
    {
        fib[i] = fib[i-1]+fib[i-2];
    }
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&n);
        int flag = 0;
        LL x,y,tz,ty;
        for(i = 45; i>2; i--)
        {
            for(ty=1; ty<=100000; ty++)
            {
                int tem = n-ty*fib[i];
                if(ty*fib[i]+fib[i-1]>n)
                    break;
                else if(tem%fib[i-1]==0&&tem/fib[i-1]<=ty)
                {
                    x = tem/fib[i-1];
                    y = ty;
                    flag = 1;
                    break;
                }
            }
            if(flag)
                break;
        }
        printf("%I64d %I64d\n",x,y);
    }

    return 0;
}