A Fi-binary number is a number that contains only 0 and 1. It does not contain any leading 0. And
also it does not contain 2 consecutive 1. The first few such number are 1, 10, 100, 101, 1000, 1001,
1010, 10000, 10001, 10010, 10100, 10101 and so on. You are given n. You have to calculate the n-th
Fi-Binary number.
Input
The first line of the input contains one integer T the number of test cases. Each test case contains one
integer n.
Output
For each test case output one line containing the n-th Fi-Binary number.
Constraints
• 1 ≤ N ≤ 109
Sample Input
4
10
20
30
40
Sample Output
10010
101010
1010001

10001001

題目大意:是Fi-binary Number的定義時由0和1構成，且沒有連續的兩個一，從第一個往後一次是 1, 10, 100, 101, 1000, 1001,
1010, 10000, 10001, 10010, 10100, 10101 。。。。。，現在給你一個n問第n項是什麼

窩找的規律麻煩了一點。。。看別人找的挺短的，，莫名uva過了，zzulioj過不了。。

我的規律是；長度相等的個數成斐波那契數，長度1的有1個，長度2的有1，長度3的有兩個，長度4的有3個。。

例如第20項為101010，先求出n屬於第幾個長度，答案是6，101010=100000+1010，,1010又等於1000+10,10是2的冪次方了（第某長度第一項。。）就結束了。。就是這個從長的往短的找。。

ac程式碼

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#define LL long long
#define INF 0x3f3f3f3f3f3f
using namespace std;
LL fac[55],sum[55];
void init()
{
    int i=1;
    fac[1]=1;
    sum[1]=1;
    sum[2]=2;
    fac[2]=1;
    for(i=3;i<=45;i++)
    {
        fac[i]=fac[i-1]+fac[i-2];
        sum[i]=sum[i-1]+fac[i];
    }
}
int bseach(LL val)
{
    int l=1,r=45;
    int ans=45;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(sum[mid]>=val)
        {
            ans=mid;
            r=mid-1;
        }
        else
            l=mid+1;
    }
    return ans;
}
LL bit[60];
LL Pow(LL a,int n)
{
    if(n==0)
        return 1;
    LL ans=1;
    int i;
    for(i=1;i<=n;i++)
        ans*=a;
    return ans;
}
int main()
{
    init();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        LL n;
        int i;
        scanf("%lld",&n);
        int c=bseach(n);
        //printf("%d\n",c);
        LL ans=0;
        ans+=Pow(2,c-1);
        while(n-sum[c-1]!=1)
        {
            n=n-sum[c-1]-1;
            //printf("%lld\n",n);
            c=bseach(n);
            ans+=Pow(2,c-1);
        }
        int k=0;
        while(ans)
        {
            LL temp=ans%2;
            bit[k++]=temp;
            ans/=2;
        }
        for(i=k-1;i>=0;i--)
            printf("%lld",bit[i]);
        printf("\n");
    }
}