hdu 1250 Hat's Fibonacci

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

题目描述：

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.

题目分析：

二维数组，但是如果每个位置存储10（即0~9中的某一个）会超内存，所以在这里我每个位置存储8位数（即0~100000000）

AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>#include<cstdlib>#define MOD 100000000using namespace std;int v[8000][305];void Init(){     v[1][1]=1;     v[2][1]=1;     v[3][1]=1;     v[4][1]=1;     for(int i=5;i<8000;i++)     {         for(int j=1;j<300;j++)            v[i][j]=v[i-1][j]+v[i-2][j]+v[i-3][j]+v[i-4][j];         for(int j=1;j<300;j++)         {             if(v[i][j]/MOD)             {                 v[i][j+1]+=(v[i][j]/MOD);                 v[i][j]%=MOD;             }         }     }}int main(){    Init();    int n;    while(scanf("%d",&n)!=EOF)    {        int k;        for(k=300;k>=1;k--)        {            if(v[n][k])                break;        }        printf("%d",v[n][k]);        for(int j=k-1;j>=1;j--)        {            printf("%08d",v[n][j]);        }        printf("\n");    }    return 0;}