题目3 : Fibonacci

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Given a sequence {an}, how many non-empty sub-sequence of it is a prefix of fibonacci sequence.

A sub-sequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

The fibonacci sequence is defined as below:

F1 = 1, F2 = 1

Fn = Fn-1 + Fn-2, n>=3

输入

One line with an integer n.

Second line with n integers, indicating the sequence {an}.

For 30% of the data, n<=10.

For 60% of the data, n<=1000.

For 100% of the data, n<=1000000, 0<=ai<=100000.

输出

One line with an integer, indicating the answer modulo 1,000,000,007.

样例提示

The 7 sub-sequences are:

{a2}

{a3}

{a2, a3}

{a2, a3, a4}

{a2, a3, a5}

{a2, a3, a4, a6}

{a2, a3, a5, a6}

 

样例输入

6
2 1 1 2 2 3

样例输出

7

Emacs Normal Vim

 

 

// Java版本
import java.awt.im.InputContext;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
/*

2
0 0
0 3

1.000 1.000 5.000

 */
    static HashSet fibSet=new HashSet();
    static    HashMap  fibMap=new HashMap();
    static    HashMap  fibMap2=new HashMap();
    public static void fib(){
        fibMap.put(1, (int) 1);
        fibMap.put(2, (int) 1);
        fibSet.add((int) 1);
        fibMap2.put((int) 1, 1);
        for(int i=3; i<50; i++){
            
            fibMap.put(i, fibMap.get(i-1)+fibMap.get(i-2));
            fibMap2.put(fibMap.get(i-1)+fibMap.get(i-2), i);
            fibSet.add(fibMap.get(i-1)+fibMap.get(i-2));
        }
        //System.out.println("fibmap");
        
    }
    //如果有a之前的所有
    public  static boolean hasPre(int a, HashMap nums){
        int k=fibMap2.get(a);
        
        boolean result=true;
        for(int i=k-1;i>2; i--){
            
            if(nums.get(fibMap.get(i)) !=null  &&nums.get(fibMap.get(i)) >0){
                continue;
            }else{
                result=false;
                break;
            }
        }
        
        if(result&& nums.get(1)!=null &&nums.get(1)>1 ){
            result= true;
        }else{
            result= false;
        }
    
        return result;
        
    }
    
    public static int precount(int a, HashMap nums){
        long count=1;
        int k=fibMap2.get(a);
    
        for(int i=k-1;i>2; i--){
            count=count*nums.get(fibMap.get(i));
                
        }
        count*=(nums.get(1)*(nums.get(1)-1)/2);
        return (int) (count%1000000007);

    }
    public static void main(String[] args) {
        
        Scanner scanner = new Scanner(System.in);
        int n=scanner.nextInt();
        int a[] = new int[n];
        for(int i=0; i){
            a[i]=scanner.nextInt();
            
        }
        fib();
        //fib计数
        HashMap  nums=new HashMap();
        boolean has1=false;
        long count=0;
        for(int i=0; ii){
            //如果等于1,那就好弄
            //System.out.println(a[i]+""+fibSet.contains((long)a[i]) );
            if(a[i]==1){
                if(nums.containsKey(1)){
                    count=count+nums.get(1)+1;
                    nums.put(1, nums.get(1)+1);
                    
                }else{
                    nums.put(1, 1);
                    count+=1;
                }
                has1=true;
                
                
                
            }else if(has1&& nums.get(1)>1&& fibSet.contains(a[i]) && hasPre(a[i],nums)){
                
                count=(count+precount(a[i],nums))%1000000007;
                if(nums.containsKey(a[i])){
                    nums.put(a[i], nums.get(a[i])+1);
                }else{
                    nums.put(a[i],  1);
                }
                
            }
            //System.out.println(nums);
        }
      
        System.out.println(count);
        scanner.close();
    }
    
}

 

转载于:https://www.cnblogs.com/stonehat/p/4847407.html