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

4203968145672990846840663646Note: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;}