HDU 1250 Hat's Fibonacci

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11104    Accepted Submission(s): 3732

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.

 

 

Author

戴帽子的

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1250

分析：这题用到大数相加，用数组的元素表示大数的各个数位的数字，（例如123，可以a[0]=3,a[1]=2,a[2]=1;）有个技巧是在网上学到的，每个数组元素存储八位数可以提高效率。先预处理，再输入数据。

下面给出AC代码：

#include<bits/stdc++.h>
using namespace std;
int a[10000][260]={0};   //每个元素可以存储8位数字，所以2005位可以用260个数组元素存储。
int main()
{
    int i,j,n;
    a[1][0]=1;      //赋初值
    a[2][0]=1;
    a[3][0]=1;
    a[4][0]=1;
    for(i=5;i<10000;i++)
    {
        for(j=0;j<260;j++)
            a[i][j]+=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];
        for(j=0;j<260;j++)                //每八位考虑进位。
            if(a[i][j]>100000000)
            {
                a[i][j+1]+=a[i][j]/100000000;
                a[i][j]=a[i][j]%100000000;
            }
    }
    while(scanf("%d",&n)!=EOF)
    {
        for(j=259;j>=0;j--)
            if(a[n][j]!=0) break;      //不输出高位的0
        printf("%d",a[n][j]);
        for(j=j-1;j>=0;j--)
            printf("%08d",a[n][j]);     //每个元素存储了八位数字，所以控制输出位数为8，左边补0
        printf("\n");
    }
    return 0;
}