SDNU 1504.B.Fibonacci

Description

Fibonacci numbers are well-known as follow:

 

 

Now given an integer N, please find out whether N can be represented as the sum of several Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers.

Input

Multiple test cases, the first line is an integer T (T<=10000), indicating the number of test cases.

 

Each test case is a line with an integer N (1<=N<=1000000000).

Output

One line per case. If the answer don’t exist, output “-1” (without quotes). Otherwise, your answer should be formatted as “N=f1+f2+…+fn”. N indicates the given number and f1, f2, … , fn indicating the Fibonacci numbers in ascending order. If there are multiple ways, you can output any of them.

Sample Input

4
5
6
7
100

Sample Output

5=5
6=1+5
7=2+5
100=3+8+89

Source

Unknown

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
const int inf = 0x3f3f3f3f;
const int maxn = 1e6;
ll number[10000+8], sum[10000+8];
int t;
ll n;
void init()
{
    for(int i = 1; i<46; i++)//虽然n<=1000000000，看起来数据很大，实际上在斐波那契数列里面，下标最大也不过是45而已
    {
        if(i == 1)number[i] = 1;
        else if(i == 2)number[i] = 2;
        else number[i] = number[i-1]+number[i-2];
    }
}
int main()
{
    while(~scanf("%d", &t))
    {
        while(t--)
        {
            init();
            int miao = 0;
            scanf("%d", &n);
            int ying = n, flag = 0;
            for(int i = 45; i>0; i--)
            {
                if(ying>=number[i])
                {
                    sum[miao++] = number[i];
                    ying -= number[i];
                    if(ying <= 0)break;
                }
            }
            printf("%d=", n);
            for(int i = miao-1; i>=0; i--)
            {
                if(flag)printf("+");
                flag = 1;
                printf("%d", sum[i]);
            }
            printf("\n");
        }
    }
    return 0;
}