hdu 1250 Hat's Fibonacci(高精度数)

//  继续大数，哎、、

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.

 

Author

戴帽子的

 

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模拟斐波那契数列 , 不过递推公式变成了 f1 = f2 = f3 = f4 = 1 ,  f(n) = f(n-1) + f(n-2) + f(n-3) +f(n-4) . (n>=5)  , 最大的数有2005位，所以，用大数吧

用的 跟 hdu1042 N! 的方法一样，用 十万 进制解决，每个整形存 10000，当然更大点也可以。。

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Code：

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int N = 7500;
const int M = 500;
int num[N][M];
void add()
{
    memset(num,0,sizeof(num));
    num[1][1] = num[2][1] = num[3][1] = num[4][1] = 1;
    int i,j,k = 0;
    for(i = 5;i<7500;i++)
        for(j = 1;j<500;j++)
        {
            k += num[i-1][j] + num[i-2][j] + num[i-3][j] + num[i-4][j];// 递推公式
            num[i][j] = k%100000;
            k = k/100000;
        }
        while(k)
        {
            num[i][j++] = k%100000;
            k = k/100000;
        }
}
int main()
{
    int n,i;
    add();
    while(cin>>n)
    {
        for(i = 500-1;i>=0;i--)
        {
            if(num[n][i]!=0)
                break;
        }
        printf("%d",num[n][i--]);
        while(i>0)
            printf("%05d",num[n][i--]);
        printf("\n");
    }
    return 0;
}