HDUOJ Hat's Fibonacci java大数处理+C++版

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13867    Accepted Submission(s): 4658

 

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.

Author

戴帽子的

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这题我用的java 写的

很无脑，

坑点一：大数数组要初始化，否则报异常

坑点二：数组要开到10000，开到7000就报错。

import java.math.BigDecimal;
import java.util.Scanner;

public class Main{

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		BigDecimal [] a=new BigDecimal[7006];
		int i,j;
		for(i=0;i<7006;i++)
		{
			if(i<4)
			    a[i]=new BigDecimal(1);
			else
				a[i]=new BigDecimal(0);
		}
		
		for(i=4;i<7006;i++)
		{
			
			a[i]=a[i].add(a[i-1]);
			a[i]=a[i].add(a[i-2]);
			a[i]=a[i].add(a[i-3]);
			a[i]=a[i].add(a[i-4]);
		}
       
		Scanner in=new Scanner(System.in);
		while(in.hasNext())
		{
			int n=in.nextInt();
			System.out.println(a[n-1]);
		}
		

	}

}

下面是别人写的C++版：

基本思路：

用一个二维数组存数列，每一个数组单元可以存4位，亦可以存1位，理论上可以存10位，因为int类型4个字节，21亿多

用一个数组存结果的位数，注意进位和最后的进位。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<string>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 7061;
const int inf = 2147483647;
const int mod = 2009;
int s[N][504],l[N];
int main()
{
    int n,i,j;
    s[1][0]=s[2][0]=s[3][0]=s[4][0]=1;
    l[1]=l[2]=l[3]=l[4]=1;      // 每一个数的位数 
    for(i=5;i<N;i++)
    {
        for(j=0;j<l[i-1];j++)
        {
            s[i][j]+=s[i-1][j]+s[i-2][j]+s[i-3][j]+s[i-4][j];
            s[i][j+1]+=s[i][j]/10000; // 处理进位 
            s[i][j]%=10000;        // 每位存4个数 
        }
        while(s[i][j]>0)           //处理最后的进位 
        {
            s[i][j+1]+=s[i][j]/10000;
            s[i][j]%=10000;j++;
        }
        l[i]=j;
    }
    while(~scanf("%d",&n))
    {
        printf("%d",s[n][l[n]-1]);
        for(i=l[n]-2;i>=0;i--)
            printf("%4.4d",s[n][i]);
        puts("");
    }
    return 0;
}

验证了一个数组单元最多能存多少位？    8位

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<string>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 7061;
const int inf = 2147483647;
const int mod = 2009;
const int M=100000000;
int s[N][504],l[N];
int main()
{
    int n,i,j;
    s[1][0]=s[2][0]=s[3][0]=s[4][0]=1;
    l[1]=l[2]=l[3]=l[4]=1;      // 每一个数的位数 
    for(i=5;i<N;i++)
    {
        for(j=0;j<l[i-1];j++)
        {
            s[i][j]+=s[i-1][j]+s[i-2][j]+s[i-3][j]+s[i-4][j];
            s[i][j+1]+=s[i][j]/M; // 处理进位 
            s[i][j]%=M;        // 每位存4个数 
        }
        while(s[i][j]>0)           //处理最后的进位 
        {
            s[i][j+1]+=s[i][j]/M;
            s[i][j]%=M;j++;
        }
        l[i]=j;
    }
    while(~scanf("%d",&n))
    {
        printf("%d",s[n][l[n]-1]);
        for(i=l[n]-2;i>=0;i--)
            printf("%8.8d",s[n][i]);
        printf("\n");
    }
    return 0;
}