1410221224-hd-Hat's Fibonacci

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7857    Accepted Submission(s): 2553

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.

题目大意

       这是斐波那契数的拓展。

解题思路

       这一方面用到了斐波那契数列，另一方面又需要大数加减法。

错误原因

       数组过大超过内存。可以将数组形式改为char型，这样可以减少四分之一的内存。

代码

#include<stdio.h>#include<string.h>char num[10000][2010];//用int型会超出内存，所以该用char型 int len[10000];//记录每个的长度 int main(){int n;int i,j,k;int a,b;for(i=1;i<10000;i++)    for(j=0;j<2010;j++)        num[i][j]='0';//初始化 num[1][0]='1';num[2][0]='1';num[3][0]='1';num[4][0]='1';len[1]=len[2]=len[3]=len[4]=1;a=0;for(i=5;i<10000;i++){b=0;for(j=0;j<=len[i-1];j++){b=a+num[i-1][j]-'0'+num[i-2][j]-'0'+num[i-3][j]-'0'+num[i-4][j]-'0';num[i][j]=b%10+'0';a=b/10;}//大数加减法 if(num[i][len[i-1]]!='0')    len[i]=len[i-1]+1;else    len[i]=len[i-1];if(len[i]>2005)    break;}while(scanf("%d",&n)!=EOF){for(i=len[n]-1;i>=0;i--)    printf("%c",num[n][i]);printf("\n");}return 0;}