HDU ACM 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<=109).

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

题解：贪心求解，此题需要注意的是相邻的两个数不能选择；

#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
long long int arr[]={,,};
int main()
{

    long long int i,j,a,b[],k,m,l,kk;
    for(i=;i<=;i++)
        arr[i]=arr[i-]+arr[i-];//先打下表
        long long int count=;
        while(scanf("%lld",&m)!=-)
        {
        for(kk=;kk<m;kk++){
            scanf("%lld",&a);
        k=;
        l=;count=;int pp=;
        for(i=;i>=;i-=)
        {
            pp=;
            count+=arr[i];
            if(count>a)
            {
                count-=arr[i];
                pp=;
            }
            else if(count==a)
            {
                k=;
                b[l++]=arr[i];
                break;
            }
            else if(count<a)
                b[l++]=arr[i];
            if(pp==)
                 i++;//判断一下上一个数是否选择，若没选，下一个数可以选择
        }
        if(k==)
        {
        printf("%d=",a);
        for(i=l-;i>=;i--)
        {

            if(i==)
                printf("%lld\n",b[i]);
            else
                printf("%lld+",b[i]);
        }
        }
        else
            printf("-1\n");
        }
        }
        return ;
}