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
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[10001]={0,1,2}; int main() { long long int i,j,a,b[123],k,m,l,kk; for(i=3;i<=90;i++) arr[i]=arr[i-1]+arr[i-2];//先打下表 long long int count=0; while(scanf("%lld",&m)!=-1) { for(kk=0;kk<m;kk++){ scanf("%lld",&a); k=0; l=0;count=0;int pp=0; for(i=90;i>=1;i-=2) { pp=0; count+=arr[i]; if(count>a) { count-=arr[i]; pp=1; } else if(count==a) { k=1; b[l++]=arr[i]; break; } else if(count<a) b[l++]=arr[i]; if(pp==1) 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==1) { printf("%d=",a); for(i=l-1;i>=0;i--) { if(i==0) printf("%lld\n",b[i]); else printf("%lld+",b[i]); } } else printf("-1\n"); } } return 0; }
