U - Hat's Fibonacci HDU - 1250

U - Hat's Fibonacci

HDU - 1250

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.

#include<cstdio>
#include<stack>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<iostream>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=7044; 
const int mmax=555;
const int MOD = 1e5+ 7;
int num[N][mmax];//数组保留对应位上的数字
int main()
{
    memset(num,0,sizeof(num));
    int mod=0;
    num[1][0]=num[2][0]=num[3][0]=num[4][0]=1;
    for(int i=5; i<=7040; i++)  //n的最大值可以说出来的，大概在7038之内
    {
        for(int j=0; j<500; j++)
        {
            mod+=num[i-1][j]+num[i-2][j]+num[i-3][j]+num[i-4][j];
            num[i][j]=mod%100000;
            mod/=100000;
        }
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n<5)
        {
            printf("1\n");
            continue;
        }
        int k=500;
        while(!num[n][k])k--;
        printf("%d",num[n][k--]);
        for(int i=k; i>=0; i--)
            printf("%05d",num[n][i]);
        printf("\n");
    }
    return 0;
}