HDU 1250 - Hat's Fibonacci（大数斐波那契）

Hat's Fibonacci

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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.

 

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先预处理一下，每4位放在二维数组的一个位置，一排代表一个斐波那契数

注意最后一个位置先输出，否则最后一个位置也按4位输出不足4位的时候会WA

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int f[8000][555];

int main()
{
    memset(f,0,sizeof(f));
    f[1][0]=f[2][0]=f[3][0]=f[4][0]=1;
    int temp=0;
    for(int i=5;i<8000;i++)
    {
        for(int j=0;j<555;j++)
        {
            temp+=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j];
            f[i][j]=temp%10000;
            temp/=10000;
        }
    }
    int n;
    while(~scanf("%d",&n))
    {
        int i;
        for(i=554;i>=0;i--)
        {
            if(f[n][i]!=0) break;
        }
        printf("%d",f[n][i--]);
        for(;i>=0;i--)
        {
            printf("%.4d",f[n][i]);
        }
        printf("\n");
    }
    return 0;
}