hdu 1250 Hat's Fibonacci(大数)

Hat's Fibonacci

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

4203968145672990846840663646Note:No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

ps：二维数组直接模拟大数加法，每一行表示一个数，每一行的每一个元素又可以代表n位数（n视情况任意而定），注意范围要大一点就好了

代码：

#include<stdio.h>#define N 10000+10int a[N][600+10];void Fibonacci(){    int i,j,k=0;    a[1][0]=a[2][0]=a[3][0]=a[4][0]=1;    for(i=5; i<N; i++)    {        for(j=0; j<600; j++)        {            k+=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];            a[i][j]=k%10000;            k/=10000;        }    }}int main(){    Fibonacci();    int n;    while(~scanf("%d",&n))    {        int i,j;        for(i=600; i>=0; i--)            if(a[n][i])                break;        printf("%d",a[n][i]);        for(j=i-1; j>=0; j--)            printf("%04d",a[n][j]);//注意0的输出        printf("\n");    }    return 0;}

总结：以前的想法太狭隘了，只是把二位数组的每一个单元存一个一位数，这次学到了二维数组的每一个单元可以存n位数的方法