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.

这个大家都懂，是个水题，知道用字符串作加法的这个肯定能懂，从左到右一位一位地加，题目中说答案不会超过2005个数字，而我一个int 存了8位，所以可以确定数组的第二维最多开个255就行，而1维嘛，10的2006次方大约等于2的7000多次方，所以开个8000足够

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>int a[8000][255]={{0}};int main(){    for(int i=1;i<=4;i++)    {        a[i][1]=1;    }    for(int i=5;i<8000;i++)    {        for(int j=1;j<255;j++)        {            a[i][j]+=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];            a[i][j+1]+=a[i][j]/100000000;            a[i][j]=a[i][j]%100000000;        }    }    int n;    while(scanf("%d",&n)!=-1)    {        int i;      for(i=254;i>=1;i--)      {          if(a[n][i]>0)          {              break;          }      }      printf("%d",a[n][i]);      for(int j=i-1;j>=1;j--)      {          printf("%.8d",a[n][j]);      }      printf("\n");    }}