HDUOJ----1250 Hat's Fibonacci

 

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5800    Accepted Submission(s): 1926

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

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

 

 

Author

戴帽子的

 

 

Recommend

Ignatius.L

  http://acm.hdu.edu.cn/showproblem.php?pid=1250

 

 大数题，我先是用打表的方法，发现数字到达7037的时候，数字规模才能达到2005 ...然后用7037*2005 ---->已经超过内存，所以这条路实不可取的。。。唯一的方法，只好暴力法，但又担心会超时，可是，喜感就是，竟然过了，还应458ms。。。。哎!！！。

贴下代码：

 1  1 #include<stdio.h>

 2  2 #include<string.h>

 3  3 #define maxn 2006    

 4  4 int a[maxn],b[maxn],c[maxn],d[maxn],ans[maxn];

 5  5 int main()

 6  6 {

 7  7     int i,j,e,s,up,num;

 8  8     while(scanf("%d",&num)!=EOF)

 9  9     {

10 10          if(num<=4)

11 11          printf("1");

12 12         else

13 13         {

14 14         memset(a,0,sizeof a);

15 15         memset(b,0,sizeof b);

16 16         memset(c,0,sizeof c);

17 17         memset(d,0,sizeof d);

18 18         memset(ans,0,sizeof ans);

19 19         a[0]=b[0]=c[0]=d[0]=1;

20 20      for(i=4,up=1;i<num;i++)

21 21      {

22 22         for(j=0,e=0;j<=up;j++)

23 23         {

24 24            s=a[j]+b[j]+c[j]+d[j]+e;

25 25            ans[j]=s%10;

26 26            e=(s-ans[j])/10;

27 27            if(s>9&&j==up) up++;

28 28            a[j]=b[j];

29 29            b[j]=c[j];

30 30            c[j]=d[j];

31 31            d[j]=ans[j];

32 32         }

33 33      }

34 34         for(j=maxn;ans[j]==0;j--);

35 35         for(i=j;i>=0;i--)

36 36         {

37 37          printf("%d",ans[i]);

38 38         }

39 39      }

40 40         printf("\n");

41 41     }

42 42 return 0;

43 43 }View Code 

View Code