HDU 1250 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.

 

思路：由于答案不会超过2005位，二维可以开个2010，但是明显过大，可以考虑优化，一个int保存8位，则只需开260即可，对于一维，10^2005大约F（10000）就可达到，故一维10000即可；

代码：

#include <iostream>#include <cstdio>using namespace std;int a[10000][260]= {0};   //每个元素可以存储8位数字，所以2005位可以用260个数组元素存储。  void init(){   int i,j;   a[1][0]=1;     //赋初值     a[2][0]=1;   a[3][0]=1;   a[4][0]=1;   for(i=5; i<10000; i++)   {       for(j=0; j<260; j++)           a[i][j]=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];       for(j=0; j<260; j++)     //每八位考虑进位             if(a[i][j]>100000000)                  {               a[i][j+1]+=a[i][j]/100000000;               a[i][j]=a[i][j]%100000000;           }    }}int main(){   int n,i,j;   init();   while(~scanf("%d",&n))   {       for(i=259; i>=0; i--)           if(a[n][i]!=0)      //不输出高位的0               break;       printf("%d",a[n][i]);       for(j=i-1; j>=0; j--)           printf("%08d",a[n][j]);    //每个元素存储了八位数字，所以控制输出位数为8，左边补0       printf("\n");    }    return 0;}