【POI2012】【BZOJ2796】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
HINT

Source

这种题都是一样的,考虑递归然后Fib数分解就行了
只不过多了个减号

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<map>#define LL long long#define MAXINT 1LL<<62using namespace std;int T;LL n,fib[100]={1,1};map<LL,int> f;int get_ans(LL x){    if (f[x])   return f[x];    int pos=lower_bound(fib,fib+88,x)-fib;    if (fib[pos]==x)    return 1;    return f[x]=min(get_ans(x-fib[pos-1]),get_ans(fib[pos]-x))+1;}int main(){    scanf("%d",&T);    for (int i=2;i<=89;i++) fib[i]=fib[i-1]+fib[i-2];    while (T--) cin>>n,printf("%d\n",get_ans(n));}