(二维数组 亿进制 或 滚动数组） Hat's Fibonacci hdu1250

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12284    Accepted Submission(s): 4124

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.

用string会超时，以下为超时代码。

#include <iostream>
#include <string>
using namespace std;
string add(string a,string b)
{
    int len1=a.length();
    int len2=b.length();
    int i;
    if(len1>len2)
    {
        for(i=1;i<=len1-len2;i++)
            b="0"+b;
    }
    else
    {
        for(i=1;i<=len2-len1;i++)
            a="0"+a;
    }
    string str;
    int cf=0,t;
    len1=a.length();
    for(i=len1-1;i>=0;i--)
    {
        t=a[i]-'0'+b[i]-'0'+cf;
        cf=t/10;
        t%=10;
        str=char(t+'0')+str;
    }
    if(cf!=0)
        str=char(cf+'0')+str;
    return str;
}
string fun(int n)
{
    string f[10010];
    f[1]="1";
    f[2]="1";
    f[3]="1";
    f[4]="1";
    int i;
    string a,b,c;
    for(i=5;i<=n;i++)
    {
        a=add(f[i-1],f[i-2]);
        b=add(f[i-3],f[i-4]);
        f[i]=add(a,b);
    }
    return f[n];
}
int main()
{
    int n;
    while(cin>>n)
    {
        string str;
        str=fun(n);
        cout<<str<<endl;
        cout<<endl;
    }
    return 0;
}

 正确的代码

方法一：
利用二维数组和亿进制。
#include<cstdio>
#include <iostream>
#include<cstring>
using namespace std; 
int str[10001][260];                //必须为int类型。
int main()
{
    memset(str,0,sizeof(str));
    str[1][0]=1;
    str[2][0]=1;
    str[3][0]=1;
    str[4][0]=1;
    int i,j,ans=0,c,n;
    for(i=5;i<10001;i++)
    {
        for(j=0,c=0;j<260;j++)          //循环，来将j个数组的8位数字的情况全部列举出。
        {
            ans=str[i-1][j]+str[i-2][j]+str[i-3][j]+str[i-4][j]+c;
            c=ans/100000000;
            str[i][j]=ans%100000000;           //每一个数组存8位数字，c来判定是否进位。
    }
    }
    while(cin>>n)
    {
        j=259;
        while(!str[n][j])       //首位有0，清除掉0。
        j--;
        cout<<str[n][j];           //开头的首0清除掉后的x位数字，可能小于8位。
        for(i=j-1;i>=0;i--)
        printf("%08d",str[n][i]);            //每8位数字输出一组，不足的自动部0。
        printf("
");
    }
    return 0;
}
方法二：
利用滚动数组求解
#include <iostream>  
#include <cstdio>  
#include <cstring>  
  
using namespace std;  
  
int t[6][2555];  
  
int main()  
{  
    int n;  
    while(scanf("%d",&n) != EOF)  
    {  
        memset(t,0,sizeof(t));  
        t[0][0] = 1;  
        t[1][0] = 1;  
        t[2][0] = 1;  
        t[3][0] = 1;  
        for(int i = 4;i < n;i++)  
        {  
            int carry = 0;  
            int k = i % 4;  
            for(int j = 0;j < 2500;j++)  
            {  
                int x = t[0][j] + t[1][j] + t[2][j] + t[3][j];  
                t[k][j] = x + carry;  
                carry = t[k][j] / 10;  
                t[k][j] %= 10;  
            }  
        }  
        int k = 2500;  
        while(t[(n - 1) % 4][--k] == 0);  
        for(int i = k;i >= 0;i--)  
        {  
            printf("%d",t[(n - 1) % 4][i]);  
        }  
          
        printf("
");  
          
    }  
      
      
    return 0;  
}

 用JAVA

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner (System.in);
        BigInteger a[]=new BigInteger[10001];
        int n;
        while(in.hasNextInt()) {
            n=in.nextInt();
            a[1]=BigInteger.ONE;
            a[2]=BigInteger.ONE;
            a[3]=BigInteger.ONE;
            a[4]=BigInteger.ONE;
            for(int i=5;i<=10000;i++) {
                a[i]=BigInteger.ZERO;
                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]);
            }
            System.out.println(a[n]);
        }
    }
}