poj 3070 Fibonacci---(矩阵快速幂模板)

Time Limit: 1000MS      Memory Limit: 65536K

矩阵快速幂

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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:

题意：

Fibonacci数列，问你第n个是多少？

解题思路：

由题+图可知：裸的矩阵快速幂，再取个模；模板模板，一切都是模板。。。。。

完整代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int mod=10000;
struct node
{
    int a[2][2];
    void init()
    {
        a[0][0]=0,a[0][1]=1;
        a[1][0]=1,a[1][1]=1;
    }
};
node matrixmul(node x,node y)
{
    node z;
    for(int i=0;i<2;i++)///题中为2*2的矩阵
    {
        for(int j=0;j<2;j++)
        {
            z.a[i][j]=0;
            for(int k=0;k<2;k++)
                z.a[i][j]+=(x.a[i][k]*y.a[k][j]);///矩阵乘法
            z.a[i][j]%=mod;
        }
    }
    return z;
}
node mul(node s,int k)
{
    node ans;
    ans.init();
    while(k>=1)
    {
        if(k&1)
            ans=matrixmul(ans,s);
        k>>=1;
        s=matrixmul(s,s);
    }
    return ans;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int ans;
        if(n==-1)
            break;
        if(n==0)
            ans=0;
        else
        {
            node s;
            s.init();
            s=mul(s,n-1);
            ans=s.a[0][1]%mod;
        }
        printf("%d\n",ans);
    }
    return 0;
}