HDU 1250Hat's Fibonacci(两种方法处理大数)

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5891    Accepted Submission(s): 1944

Problem Description

A Fibonacci sequence is calculated by adding the prevIoUs two members the sequence,with the first two members being both 1.
F(1) = 1,F(2) = 1,F(3) = 1,F(4) = 1,F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input,and print that Fibonacci number.

 

Input

Each line will contain an integers. Process to end of file.

 

Output

For each case,output the result in a line.

 

Sample Input

  
  

   
   100
  
  

 

Sample Output

  
  

   
   4203968145672990846840663646


Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data,ie. F(20) = 66526 has 5 digits.
  
  

 

                   题目大意：题目的意思就不说了，主要是简单地处理大数。

            解题思路：先是用C++写的处理大数，每次只涉及到两个数相加，直接转换成string，所以位数就取len较大的那位。如果第一位超过了10则直接再最前面加一个1，第一位减去10，string支持直接用"1"+。详见代码。

            题目地址：Hat's Fibonacci

C++代码：

#include<iostream>   
#include<string>    
using namespace std;  

string add(string s1,string s2)  
{  
      
    int j,l,la,lb;  
    string ma,mi;  
    ma=s1;mi=s2;  
    if(s1.length()<s2.length())
        {ma=s2;mi=s1;}  
    la=ma.size();lb=mi.size();  
    l=la-1;  
    for(j=lb-1;j>=0;j--,l--) 
        ma[l] += mi[j]-'0';   
    for(j=la-1;j>=1;j--) 
        if(ma[j]>'9')
            {ma[j]-=10;ma[j-1]++;}  
    if(ma[0]>'9')   //处理第一位超过9了。
        {ma[0]-=10;ma='1'+ma;}  
    return ma;  
}  


int main(){  
    int n,i;  
    string a[8008];  
    a[1]="1";  
    a[2]="1";  
    a[3]="1";  
    a[4]="1";  
    for(i=5;i<8008;++i)  
       a[i]=add(add(add(a[i-1],a[i-2]),a[i-3]),a[i-4]);  
    while(cin>>n)  
        cout<<a[n]<<endl;  

    return 0;       
}  

          以后遇到大数的问题，不用那么坑爹的再敲该死的那么长的模板了，直接上JAVA。

JAVA代码：

import java.util.*;
import java.math.*;

public class Main {
     public static void main(String args[])
     {
    	   Scanner cin = new Scanner(system.in);
    	   BigInteger res[];
    	   res = new BigInteger[8008];
    	   res[1]=BigInteger.ONE;
    	   res[2]=BigInteger.ONE;
    	   res[3]=BigInteger.ONE;
    	   res[4]=BigInteger.ONE;
    	   for(int i=5;i<=8000;i++)
    	   {
    		   res[i]=res[i-1].add(res[i-2]);
    		   res[i]=res[i].add(res[i-3]);
    		   res[i]=res[i].add(res[i-4]);
    	   }
    	   int p;
    	   while(cin.hasNext())
    	   {
    		   p=cin.nextInt();
    		   System.out.println(res[p]);
    	   }
     }
}