v4t3561(拆斐波那契数)

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博客围绕Fibonacci数展开，给定整数N，需判断其能否表示为若干不相邻Fibonacci数之和。介绍了输入输出格式，包含多组测试用例，给出示例。还提示只需存小于1e9的数，从大向小遍历求解。

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<=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.

Sample

Input

Copy4
5
6
7
100

Output

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

不需要用long long,只存小于1e9的数，然后从大向小遍历。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int i,t,n,a[50],b[50];
    a[1]=1;
    a[2]=2;
    for(i=3;i<=45;i++)
        a[i]=a[i-1]+a[i-2];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int cnt=0,k=n;
        for(i=45;i>0;i--)
        {
            if(a[i]<=n)
            {
                b[cnt++]=a[i];
                n=n-a[i];
            }
        }
        if(!n)
        {
            cout<<k<<"=";
            for(i=cnt-1;i>=0;i--)
            {
                if(i<cnt-1)
                    cout<<"+";
                cout<<b[i];
            }
            cout<<'\n';
        }
        else
            cout<<"-1"<<'\n';
    }
    return 0;
}