Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

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.

/************************************************************************/

题意：首先，该题给出了一个类似于斐波那契数列的数列的前4项以及它的递推式，要求我们求出第n项(关于n的范围，题目只告诉我们任何一项的值位数都在2005位以内)

解题思路：因为递推式也已经知道了，此题无非就是一道大数加法的题目，解决方案大体分为两种

①对于给出的n，每次都从第5项开始重新计算，这样做节省了空间，但是时间复杂度就比较高了，几百MS应该是要的吧

②另一种则无非是空间换时间，在预处理阶段就先提前算好所有项的值

第②方法15MS就差不多了，但是我们可以来算算，每一项至少要留出2005的空间，而根据测试到 7060项的时候长度为 2012位，那就至少得开一个7061*2012的字符数组

orz

难逃MLE的命运

然而，这里有一种甚妙的解决方法：

我们都清楚一点，平时模拟大数加法的时候，我们都是数组的1个单元表示数的一位，比如说1314，我们或许就开了s[4]的数组，每一下标地址存储1314的一位，其实，我们完全可以只开s[1]的数组，存取1314的四位，这样就节省了3/4的空间，当然，你也可以存5位或者6位等等

以下给出该数列一些项的值以供参考：

第1项:
1
第2项:
1
第3项:
1
第4项:
1
第5项:
4
第6项:
7
第7项:
13
第8项:
25
第9项:
49
……
第20项:
66526
……
第100项:
4203968145672990846840663646
……
第7060项:
19132815656249268518490884190898848150001617262347102273710143150756354315829469888479509852889801050570395030799042735599111306649112251672457076466650962498630829437743787496112628531755305195115824430566997259641082744720736373287527919747575700260158473417665109405226920319652777277465175585464477696741922610415720562637391000667082785866914124869954720605846206351934256829531337834907474790842402939533274168300465378314603690508388596785309644903523761912630349267978975515647065125864382428117352761163937139356224263020486240923594734374261980140548731752455902881133635264737507924434810945115433313620016890039236095984940057274493964779264477753883735193438337115328806060914311429064446374554426902401691918403227167000416965835533574488228673331213305584089827205822198178813616407932298992968043206307140383955468912432121406775274033553963870186469871676404048161402129097748941437460852413334088109674329206846676321465777732569214238064950886012062268873206493753190634789004799611556800637203227512723105856795530816513091583725538824859376293542708875845281047618923884839453333724411835902641527124883119228499146282120886161631896324333116263087312981248089803618917481853788515567591309175795141248776102795891234197235862536601235399047369123830729712393846351332113738084379456593883504192488947323489128239861181670430489724151486093702031373256044063969823485993332685133721886784746661986430213891375782817130484592345089812001975466681122284637215311187205376701799534964335815653819205997139763191833257419798477962899189856200584555253213674148433904217604249741835039321036538135988714461282791837758943144139298372850687703481995710114234187808790725517079076758858226460912581124809100937111136550336603385642706799893940658422716366496495750738210734489036386052479471041621146201924331588144572267285937979746436701951413929888154167243166012017529783432709278881378267793177030592421107611791666043469274080669610510569035691019602251413731655081987038208970322896071458782

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 7061;
const int inf = 2147483647;
const int mod = 2009;
int s[N][504],l[N];
int main()
{
    int n,i,j;
    s[1][0]=s[2][0]=s[3][0]=s[4][0]=1;
    l[1]=l[2]=l[3]=l[4]=1;
    for(i=5;i<N;i++)
    {
        for(j=0;j<l[i-1];j++)
        {
            s[i][j]+=s[i-1][j]+s[i-2][j]+s[i-3][j]+s[i-4][j];
            s[i][j+1]+=s[i][j]/10000;
            s[i][j]%=10000;
        }
        while(s[i][j]>0)
        {
            s[i][j+1]+=s[i][j]/10000;
            s[i][j]%=10000;j++;
        }
        l[i]=j;
    }
    while(~scanf("%d",&n))
    {
        printf("%d",s[n][l[n]-1]);
        for(i=l[n]-2;i>=0;i--)
            printf("%4.4d",s[n][i]);
        puts("");
    }
    return 0;
}

菜鸟成长记

HDU 1250 Hat's Fibonacci(递推+大数加法)

Hat's Fibonacci

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