【bzoj2796】 [Poi2012]Fibonacci Representation

【bzoj2796】 [Poi2012]Fibonacci Representation

Description

Fib数列0,1,1,2,3,5,8,13,21。

给出一个数字，用FIB数列各项加加减减来得到。例如

10=5+5

19=21-2

17=13+5-1

1070=987+89-5-1

Input

In the first line of the standard input a single positive integer is given （1 <=P<=10） that denotes the number of queries. The following lines hold a single positive integer K each 1<=K<=10^17.

Output

For each query your program should print on the standard output the minimum number of Fibonacci numbers needed to represent the number k as their sum or difference.

Sample Input

1
1070

Sample Output

4

题解

因为F[k]*2=F[k+1]+F[k-2]，即存在最优解满足同一个FIB数出现次数不超过1

令l表示小等于n且最大的斐波那契数，r为其后一项，可以暴力或者二分求

dp(n) = min{ dp(n-l) , dp(r-n) } + 1

使用记忆化搜索

 

#include<set>
#include<map>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>

#define ll long long
#define inf 2e18
using namespace std;
ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

int T,top;
ll n,f[1005];
map<ll,int>F;

int solve(ll n)
{
    if(F[n])return F[n];
    int b=lower_bound(f,f+top,n)-f;
    if(f[b]==n)return 1;
    return F[n]=min(solve(n-f[b-1]),solve(f[b]-n))+1;
}
int main()
{
    f[0]=1;f[1]=1;
    for(int i=2;f[i-1]<=inf;i++,top++)
        f[i]=f[i-1]+f[i-2];
    T=read();
    while(T--)
    {
        n=read();
        printf("%d\n",solve(n));
    }
}