poj 3070 Fibonacci 矩阵相乘

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7715		Accepted: 5474

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

Source

Stanford Local 2006

 

想学高斯消元，先学矩阵。

二分思想，在A^B mod m中已经体现..

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>

struct node
{
    int a[5][5];
}now,cur;


void make_first()
{
    now.a[1][1]=1;
    now.a[1][2]=1;
    now.a[2][1]=1;
    now.a[2][2]=0;

    cur.a[1][1]=1;
    cur.a[1][2]=1;
    cur.a[2][1]=1;
    cur.a[2][2]=0;
}

struct node make_cheng(node  a,node b) //发现结构体忘记了。
{
    struct node f;
    int i,k,j;
    memset(f.a,0,sizeof(f.a));
    for(i=1;i<=2;i++)
    for(k=1;k<=2;k++)
    if(a.a[i][k])
    {
        for(j=1;j<=2;j++)
        if(b.a[k][j])
        {
            f.a[i][j]+=a.a[i][k]*b.a[k][j];
            if(f.a[i][j]>=10000)
            f.a[i][j]%=10000;
        }
    }
    return f;
}

void make_EF(int n)
{
    make_first();
    while(n)
    {
        if(n&1)
        {
            now=make_cheng(now,cur);//！！！
        }
        n=n/2;
        cur=make_cheng(cur,cur);
    }
    printf("%d
",now.a[1][1]);
}

int main()
{
    int n;
    while(scanf("%d",&n)>0)
    {
        if(n==-1)break;
        if(n==0){printf("0
");continue;}
        if(n==1||n==2){printf("1
");continue;}
        make_EF(n-2);
    }
    return 0;
}