Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 268 Accepted Submission(s): 112

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

Sample Output

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

用二维数组模拟大数加法，每一行表述一个数，每一行的一个元素可以代表n位数，这个可以根据自己的需要自己定义。

其他的就和正常的加法一样了，注意进位处理。

#include <iostream>

#include <stdio.h>

using namespace std;
int s[7500][670];
void solve(){

s[1][1] = 1;s[2][1] = 1;

    s[3][1] = 1;s[4][1] = 1;
    int i,j,k=0;
    for(i = 5;i<7500;i++)
    for( j = 1;j<=670 ;j++)
    {

k += s[i-1][j]+s[i-2][j]+s[i-3][j]+s[i-4][j];

s[i][j] = k%10000;

        k = k/10000;
    }
    while(k)
    {

s[i][j++] = k%10000;

        k = k/10000;
    }
}
int main()
{

    int n,i,j;

    solve();
    while(cin>>n)
    {

        for(i = 670; i>=1;i--)
        if(s[n][i]!=0)break;

printf("%d",s[n][i]);
for(j = i-1;j>=1;j--)

printf("%04d",s[n][j]);

printf("\n");
}
}

本文转自NewPanderKing51CTO博客，原文链接：

http://www.cnblogs.com/newpanderking/archive/2011/07/31/2122528.html

，如需转载请自行联系原作者