HDU1250(大数相加)

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4998    Accepted Submission(s): 1691

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.
  
  

 

///运用数组存储固定邻接位数

#include<iostream> #include<iomanip> using namespace std; struct {  int link[600];  int num; }a[10010]; int main() {  int n,i,j,temp;  a[1].link[a[1].num++]=1;  a[2].link[a[2].num++]=1;  a[3].link[a[3].num++]=1;  a[4].link[a[4].num++]=1;  for(i=5;i<=10000;i++)  {   temp=0;   for(j=0;j<=a[i-1].num-1;j++)   {    a[i].link[j]=a[i-1].link[j]+a[i-2].link[j]+a[i-3].link[j]+a[i-4].link[j]+temp;    temp=a[i].link[j]/10000;    a[i].link[j]=a[i].link[j]%10000;   }   a[i].num=a[i-1].num;   if(temp>0)   {    a[i].link[j]=temp;    a[i].num++;   }  }  while(cin>>n)  {   j=a[n].num-1;   cout<<a[n].link[j];   for(j=a[n].num-2;j>=0;j--)   {    printf("%04d",a[n].link[j]);     }   cout<<endl;   }  return 0; }