每个正整数都可以分解成斐波那契数列中的几个数相加……

从大到小贪心法就可以了……

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define sc(a,b) scanf("%d%d",&a,&b)
#define pri(a) printf("%d\n",a)
#define lson i<<1,l,mid
#define rson i<<1|1,mid+1,r
#define MM 1000005
#define MN 2005
#define INF 100004
#define eps 1e-7
using namespace std;
typedef long long ll;
string solve(int n) //n的斐波那契进制转换
{
    static int fib[305];
    fib[0]=fib[1]=1;
    int i,j;
    for(i=2;i<300;i++)
    {
        fib[i]=fib[i-1]+fib[i-2]; //斐波那契数列
        if(fib[i]>n) {j=i;break;}
    }
    string s;
    for(i=j-1;i>0;i--)
        if(fib[i]<=n) s+=‘1‘,n-=fib[i];
        else s+=‘0‘;
    return s;
}
int main()
{
    int t;
    sca(t);
    while(t--)
    {
        int n;
        sca(n);
        string s=solve(n);
        cout<<n<<" = "<<s<<" (fib)"<<endl;
    }
	return 0;
}