山东省第七届ACM省赛题——Fibonacci（二分）

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.
Example Input

4
5
6
7
100
Example Output

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

求输入的数是否满足多个斐波那契数相加，如果满足，从小到大输出
二分答案，1e9内的斐波那契数也才50个

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <math.h>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <ctime>
#define INF 0x3f3f3f3f
#define MAXN 10005
#define Mod 1000000007
using namespace std;
long long f[MAXN],cnt;
void init()
{
    f[0]=0;
    f[1]=1;
for(cnt=2; f[cnt-1]<=10000000000; ++cnt)
        f[cnt]=f[cnt-1]+f[cnt-2];
}
long long ans[100];
int main()
{
    cnt=0;
    init();
//cout<<cnt<<endl;
int t;
scanf("%d",&t);
while(t--)
    {
long long n,nn;
int p=0;
scanf("%lld",&n);
        nn=n;
while(n)
        {
int low=1,high=cnt,mid;
while(low<=high)
            {
                mid=(low+high)>>1;
if(f[mid]==n)
                {
                    high=mid;
break;
                }
else if(f[mid]<n)
                    low=mid+1;
else
                    high=mid-1;
            }
            n-=f[high];
            ans[p++]=f[high];
        }
printf("%lld=%lld",nn,ans[p-1]);
for(int i=p-2;i>=0;--i)
printf("+%lld",ans[i]);
printf("\n");
    }
return 0;
}