hdu 1250

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5445    Accepted Submission(s): 1820

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.

 

#include<stdio.h>
#include<string.h> 
int a[8000][300];
int main() 
{ 
    int i,j,n,num[8000];
    memset(a,0,sizeof(a));
    memset(num,0,sizeof(num));
    a[1][0]=a[2][0]=a[3][0]=a[4][0]=1; 
    num[1]=num[2]=num[3]=num[4]=1; 
    for(i=5;i<8000;i++)  //打表
    { 
        for(j=0;j<num[i-1];j++)  //num[]用来储存a[i][j]的位数
        { 
            a[i][j]+=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j]; 
            a[i][j+1]+=a[i][j]/100000000; 
            a[i][j]%=100000000; 
        } 
        num[i]=num[i-1]; 
        if(a[i][num[i]]!=0)
            num[i]++; 
    } 
    while(~scanf("%d",&n)) 
    { 
        if(n==0) 
        {
            printf("0\n");
            continue;
        } 
     printf("%d",a[n][num[n]-1]); 
        for(i=num[n]-2;i>=0;i--)
            printf("%08d",a[n][i]);  //位数不足8位的，在前边补充0
        printf("\n"); 
    } 
    return 0; 
}