山东省第七届ACM大学生程序设计竞赛-Fibonacci

Fibonacci

Time Limit: 2000ms   Memory limit: 131072K  有疑问？点这里^_^

题目描述

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.

输入

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).

输出

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.

示例输入

4

5

6

7

100

示例输出

5=5

6=1+5

7=2+5

100=3+8+89

提示

 

来源

 “浪潮杯”山东省第七届ACM大学生程序设计竞赛

题目意思：

给一个数N，求出是否有互不相邻的几个斐波那契数相加等于N。

解题思路：

先找到满足小于等于n的最大斐波那契数，再依次递减寻找下一个满足条件的数。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define MAXN 1001
int f[50],a[50];//分别表示保存斐波那契数列、求和的各项
void fun()//斐波那契数列打表
{
f[0]=1,f[1]=2;
for(int i=2; i<50; ++i)
f[i]=f[i-1]+f[i-2];
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t,flag,cnt;
cin>>t;
fun();
memset(a,0,sizeof(a));
while(t--)
{
int i,j,n;
flag=cnt=0;
cin>>n;
for(i=0; i<50; ++i)//找到满足小于等于n的最大斐波那契数
if(n<=f[i])
{
cout<<n<<"=";
if(n==f[i]) a[cnt++]=f[i];
else
{
--i;
a[cnt++]=f[i];
}
flag=1;//flag标记是否存在求和序列
n-=f[i];
break;
}
for(j=i-1; j>=0; --j)//i-1是因为不能相邻
{
if(n>=f[j])
{
n-=f[j];
a[cnt++]=f[j];
--j;
}
}
if(!flag) cout<<"-1"<<endl;//不存在
else
{
for(i=--cnt; i>0; --i)
cout<<a[i]<<"+";
cout<<a[0]<<endl;
}
}
return 0;
}
/*
4
5
6
7
100
*/