hdoj 1250 <10000以内Fibonacci---数组打表>

本文链接：https://blog.youkuaiyun.com/leibniz_zhang/article/details/52061404

本文介绍了一个计算特殊四阶斐波那契数列的C++程序实现，该数列通过累加前四项来生成后续项，并针对大数处理进行了优化。

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Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10566 Accepted Submission(s): 3506

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

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

内存狂爆--每个数组最大可以存8个数--

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int shu[10050][420];
int main()
{
    memset(shu,0,sizeof(shu));
    shu[1][1]=shu[2][1]=shu[3][1]=shu[4][1]=1;
    for (int i=5;i<=10020;i++)
    {
        for (int j=1;j<=410;j++)
        {
            shu[i][j]+=shu[i-1][j]+shu[i-2][j]+shu[i-3][j]+shu[i-4][j];
            if (shu[i][j]>=10000000)
            {
                shu[i][j+1]=shu[i][j]/10000000;
                shu[i][j]%=10000000;
            }
        }
    }
    int n;
    while (~scanf("%d",&n))
    {
        if (n<5)
            printf("1\n");
        else
        {
            bool fafe=false;
            for (int i=410;i>0;i--)
            {
                if (fafe)
                {
                    printf("%07d",shu[n][i]);
                }
                else if (shu[n][i])
                {
                	fafe=true;
                    printf("%d",shu[n][i]);
				}
            }
            printf("\n");
        }
    }
    return 0;
}