hdu1250 Hat's Fibonacci（大数加法二）模板

 

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3442    Accepted Submission(s): 1169

 

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.

 

分析：

大数加法一

因为有2005位，大概估计一下第8000项可以超过2005位 ，时间复杂度是8000*2000，也就是O（10^7）

#include<iostream>
#include<string>
#define N 7050
using namespace std;


string bigmult(string a,string b)
{
    int la,lb,lc,i,j,flag;
    string c="";
    char t;
    i=la=a.length()-1;
    j=lb=b.length()-1;
    flag=0;
    while(i>=0&&j>=0)
    {
        t=a[i]+b[j]+flag-'0';
        if(t>'9') 
        {
            t=t-10;flag=1;
        }
        else flag=0;
        c+=t;
        i--;j--;
    }
    while(i>=0)
    {
        t=a[i]+flag;
        if(t>'9')
        {
            t=t-10;flag=1;
        }
        else flag=0;
        c+=t;
        i--;
    }
    while(j>=0)
    {
        t=b[j]+flag;
        if(t>'9')
        {
            t=t-10;flag=1;
        }
        else flag=0;
        c+=t;
        j--;
    }
    if(flag) c+=(flag+'0');
    lc=c.length();
    for(i=0,j=lc-1;i<j;i++,j--)
    {
        t=c[i];c[i]=c[j];c[j]=t;
    }
    return c;
}

string f[N];

int main()
{
    int n,i;
    f[1]=f[2]=f[3]=f[4]="1";
    for(i=5;i<N;i++)
    {
        f[i]=bigmult(f[i-1],f[i-2]);
        f[i]=bigmult(f[i],f[i-3]);
        f[i]=bigmult(f[i],f[i-4]);
    }
    while(cin>>n)
    {
        cout<<f[n]<<endl;
    }
    return 0;
}

 

对大数加法一改进

对大数的加法一段一段的加，也就是没 8 位看成一个整形的数（mod 10^9），进行整数加法；

时间复杂度8000*2000/8 ,也就是 O（10^6）

如果改用int64时间复杂度还可以降下去。

#include<iostream>
#include<cstring>
#define N 10000
#define M 300
using namespace std;

int f[N][M];

void work()
{
    memset(f,0,sizeof(f));
    f[1][1]=f[2][1]=f[3][1]=f[4][1]=1;
    int i,j,t;
    for(i=5;i<N;i++)
    {
        t=0;
        for(j=1;j<M;j++)
        {
            t=t+f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j];
            f[i][j]=t%100000000;
            t=t/100000000;
        }
    }
}

int main()
{
    int n,i;
    work();
    while(scanf("%d",&n)!=EOF)
    {
        i=M-1;
        while(f[n][i]==0)i--;
        printf("%d",f[n][i--]);
        while(i>=1)
        {
            printf("%08d",f[n][i]);
            i--;
        }
        printf("\n");
    }
    return 0;
}