hdu 1250 Hat's Fibonacci(高精度加法)

题目：​​http://acm.hdu.edu.cn/showproblem.php?pid=1250​​

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.

2005位的话用高精度计算，算到第8000多应该就够了。

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=2008;
char f[9000][maxn+2];      //为什么取9000？
int main()
{
    int i=5,p=maxn,n,num;
    f[1][maxn]=f[2][maxn]=f[3][maxn]=f[4][maxn]=1;
    while(f[i-1][1]<=1){
         for(int j=maxn;j>=p;j--){
              f[i][j]=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j];  //******
         }
         for(int j=maxn;j>=p;j--){
              int c=f[i][j]/10;
              if(c>0){
                  f[i][j]=f[i][j]%10;
                  f[i][j-1]+=c;
              }
         }
         if(f[i][p-1]>0)p--;
         i++;
    }
    while(cin>>n){
         for(int k=0;k<=maxn;k++){
              if(f[n][k]!=0){
                  num=k;
                  break;
              }
         }
         for(int k=num;k<=maxn;k++)printf("%d",f[n][k]);
         puts("");
    }
    return 0;
}