2016山东ACM省赛B 3561Fibonacci_2016山东acm竞赛b-优快云博客

本文介绍了一个算法问题，即如何判断一个整数是否可以表示为若干个不相邻的斐波那契数之和，并提供了一段C++代码实现。通过预计算斐波那契数列并逆向减法的方式进行判断。

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Fibonacci

Problem 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<=10^9).

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.

Example Input

Example Output

5=5

6=1+5

7=2+5

100=3+8+89

#include <iostream>
using namespace std;
int f[45]={1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,165580141,267914296,433494437,701408733,1134903170};
//斐波那契数列打表到10^9，递归求的话可能会超时

int a[50];//存放加数

int main()
{
    int t,n,k;
    cin>>t;
    while(t--)
    {
        cin>>n;
        k=n;
        int flag=0;
        for(int i=45;i>=1;i--)
        {
            if(f[i]<=n)
            {
                a[flag++]=f[i];
                n=n-f[i];
            }
        }
        if(n==0)
        {
            cout<<k<<"=";
            for(int i=flag-1;i>=1;i--)
            {
                if(i==flag-1)cout<<a[i];  //第一个
                else cout<<"+"<<a[i];
            }
            cout<<endl;
        }
        else cout<<"-1"<<endl;
    }
    return 0;
}