Hat's Fibonacci

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.

纪念构建大数模板成功！！！！！（大一上学期写的，所以代码有点糙，献丑了0.0）

代码：

#include

using namespace std;

vectorv;

string big_num(string nl,string ml)

{

    //cout<<nl<<" "<<ml<<endl;

    string str="";

    int t,len,lon,j,x=1,l,m[2050],n[2050];

    memset(m,0,sizeof m);

    memset(n,0,sizeof n);

    j=0;

    len=nl.size();

    lon=ml.size();

    for(int i=len-1;i>=0;i--)

    {

        n[i]=nl[j++]-'0';

        //cout<<n[i]<<endl;

    }

    j=0;

    for(int i=lon-1;i>=0;i--)

    {

        m[i]=ml[j++]-'0';

    }

    if(len

    l=lon;

    else

    l=len;

    for(int j=0;j<=l-1;j++)

    {

        n[j]+=m[j];

        if(n[j]>=10)

        {

            n[j+1]=n[j+1]+1;

            n[j]=n[j]-10;

        }

    }

    if(n[l]==0)

    {

        //cout<<n[0]<<endl;

        for(int i=l-1;i>=0;i--)

        {

            str+=(n[i]+'0');

            //cout<<str<<endl;

            //cout<<n[i];

        }

        return str;

    }

    else

    {

        for(int i=l;i>=0;i--)

        {

            str+=(n[i]+'0');

            //cout<<str<<endl;

        }

        //cout<<n<<endl;

        return str;

    }

}

void solve()

{

    v.push_back("1");

    v.push_back("1");

    v.push_back("1");

    v.push_back("1");

    v.push_back("1");

    for(int i=5;;i++)

    {

        //cout<<v[i-1]<<" "<<v[i-2]<<" "<<v[i-3]<<" "<<v[i-4]<<endl;

        v.push_back(big_num(big_num(v[i-1],v[i-2]),big_num(v[i-3],v[i-4])));

       // cout<<"前四项为：";

        //cout<<v[i-1]<<" "<<v[i-2]<<" "<<v[i-3]<<" "<<v[i-4]<<endl;

        //cout<<"和为：";

        //cout<<v[i]<<endl;

        if(v[i].size()>2006)

            return;

    }

}

int main()

{

    //freopen("in.txt", "r", stdin);

    solve();

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        //cout<<"前四项为：";

        //cout<<v[n-1]<<" "<<v[n-2]<<" "<<v[n-3]<<" "<<v[n-4]<<endl;

        //cout<<"和为：";

        cout<<v[n]<<endl;

    }

    return 0;

}