HDU1250：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.

 

 

大数模板题

 

#include <iostream>

#include <cstdio>

#include <vector>

#include <algorithm>

#include <cmath>

#include <string.h>

#include <malloc.h>

using namespace std;

void add(char* a,char* b,char* c)

{

    int i,j,k,max,min,n,temp;

    char *s,*pmax,*pmin;

    max=strlen(a);

    min=strlen(b);

    if (max<min)

    {

        temp=max;

        max=min;

        min=temp;

        pmax=b;

        pmin=a;

    }

    else

    {

        pmax=a;

        pmin=b;

    }

    s=(char*)malloc(sizeof(char)*(max+1));

    s[0]='0';

    for (i=min-1,j=max-1,k=max; i>=0; i--,j--,k--)

        s[k]=pmin[i]-'0'+pmax[j];

    for (; j>=0; j--,k--)

        s[k]=pmax[j];

    for (i=max; i>=0; i--)

        if (s[i]>'9')

        {

            s[i]-=10;

            s[i-1]++;

        }

    if (s[0]=='0')

    {

        for (i=0; i<=max; i++)

            c[i-1]=s[i];

        c[i-1]='\0';

    }

    else

    {

        for (i=0; i<=max; i++)

            c[i]=s[i];

        c[i]='\0';

    }

    free(s);

}

char a[8001][2505];

int main(void)

{

    int n,i;

    for(i=1; i<=4; i++)

        strcpy(a[i],"1");

    for(i=5; i<=8000; i++)

    {

        char c[2505],b[2505];

        add(a[i-1],a[i-2],c);

        add(a[i-3],a[i-4],b);

        add(b,c,a[i]);

    }

    while(cin>>n)

    {

        cout<<a[n]<<endl;

    }

    return 0;

}