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
<<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]);
}
}
}