hdu Hat's Fibonacci

Hat's Fibonacci

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

Total Submission(s): 789    Accepted Submission(s): 343

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

 

Recommend

Ignatius.L

分析：大整数加法，利用分部整数数组计算

#include<stdio.h>
#include<string.h>
int main()
{
    char a[6],r[201];
    int n,i,j,x,count,jw,b[6],count1;
    int s[201];

    while(scanf("%s%d",a,&n)==2)
    {
        memset(r,0,sizeof(r));
        count=x=count1=0;
        for(i=5;i>=0;--i)
        {
            if(a[i]!='.')
            {
                b[count]=a[i]-'0';
                ++count;
            }
            else
                x=n*(5-i);
        }
        for(i=0;i<x/n;++i)
        {
            if(!b[i])
            {
                ++count1;
            }
            else
                break;
        }
        x-=n*count1;
        for(i=0;i<count-count1;++i)
        {
            b[i]=b[i+count1];
            r[200-i]=b[i];
        }

        while(--n)
        {
            memset(s,0,sizeof(s));
            for(j=0;j<count-count1;++j)
            {
               for(i=200;i>=6;--i)
               {
                s[i-j]+=r[i]*b[j];
               }
            }
            jw=0;
            for(i=200;i>=0;--i)
            {
                r[i]=(s[i]+jw)%10;
                jw=(s[i]+jw)/10;
            }
        }

        if(!x)
        {
            for(i=0;i<201;++i)
                if(r[i])
                    break;
            if(i==201)
                printf("0");
            else
            {
                for(;i<201;++i)
                    putchar(r[i]+'0');
            }
        }
        else
        {
            for(i=0;i<=200-x;++i)
                if(r[i])
                    break;
            for(;i<=200-x;++i)
                putchar(r[i]+'0');
            putchar('.');
            for(;i<=200;++i)
                putchar(r[i]+'0');
        }
        putchar(10);
    }
return 0;
}