HDU 1250-Hat's Fibonacci

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4730    Accepted Submission(s): 1610

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.

Output

For each case, output the result in a line.

 

题目给出公式
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)，然后给出n，要求出F[n]，数据比较大，要用到大数运算

AC code:

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<algorithm>



using namespace std;



int f[4][2100];

int ans[2100];



int main()

{

    int N;



    while(cin>>N)

    {

        if(N<=4)

        {

            cout<<"1"<<endl;

            continue;

        }

        memset(f,0,sizeof f);

        memset(ans,0,sizeof ans);

        f[0][0]=f[1][0]=f[2][0]=f[3][0]=1;

        int len=1;

        for(int i=4;i<N;i++)

        {

            int carry=0;

            for(int j=0;j<len;j++)

            {

                int temp=carry+f[0][j]+f[1][j]+f[2][j]+f[3][j];

                ans[j]=temp%10;

                carry=temp/10;

            }

            while(carry!=0)

            {

                ans[len++]=carry%10;

                carry/=10;

            }

            for(int j=0;j<len;j++)

            {

                f[i%4][j]=ans[j];

            }

        }

        for(int i=len-1;i>=0;i--)

            cout<<ans[i];

        cout<<endl;

    }

}