HDUOJ Hat's Fibonacci java大数处理+C++版

悠悠 2024-02-19 19:34 16阅读 0赞

## 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

Recommend

Ignatius.L | We have carefully selected several similar problems for you: [1753][] [1865][] [1715][] [1002][] [2100][]

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;
    }

![20190328172602384.png][]

[1753]: http://acm.hdu.edu.cn/showproblem.php?pid=1753
[1865]: http://acm.hdu.edu.cn/showproblem.php?pid=1865
[1715]: http://acm.hdu.edu.cn/showproblem.php?pid=1715
[1002]: http://acm.hdu.edu.cn/showproblem.php?pid=1002
[2100]: http://acm.hdu.edu.cn/showproblem.php?pid=2100
[20190328172602384.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/01/29/26a21e1bf8d04a88b00e799b6a353d55.png