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.
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
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(" "); 41 41 } 42 42 return 0; 43 43 }View Code