大数变形版斐波那契

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
1**00

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[7500][600];
int main()
{
    memset(a,0,sizeof(a));
    int i,j,k,n,m;
    a[2][0]=1;
    a[1][0]=1;
    a[3][0]=1;
    a[4][0]=1;
    int g=0;
    for(i=5;i<7500;i++)
    {
        for(j=0;j<=g;j++)
        {
            a[i][j]=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];
        }
        for(j=0;j<=g;j++)
        {
            a[i][j+1]+=a[i][j]/10000;//每位存放10000,一开始存放10,结果超内存
            a[i][j]%=10000;
        }
        if(a[i][g])
        {
            g++;
        }
    }
    int h;
    while(scanf("%d",&h)!=EOF)
    {
        for(i=g;i>=0;i--)
            if(a[h][i]!=0)
                break;
            printf("%d",a[h][i--]);
        for(;i>=0;i--)
            printf("%04d",a[h][i]);//不足4位前置0
        printf("
");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/nanfenggu/p/7900193.html