【矩阵快速幂】POJ 3070 Fibonacci （大数 Fibonacci）（大二版）

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n = F_{n ? 1} + F_{n ? 2} forn ≥ 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.

【解题思路】

练习矩阵快速幂的基础题了，不说了，比较基础，初学快速幂的可以看这里 click here~~   click here~~加深理解

本题注意取模mod的大小

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;
const int mod[5]= {0,10,100,1000,10000};
const int MOD =10000;
#define LL long long
LL X,Y,N,M,i,j;
struct Matrlc
{
    int  mapp[2][2];
} ans,base;
Matrlc unit= {1,0,0,1};
Matrlc mult(Matrlc a,Matrlc b)
{
    Matrlc c;
    for(int i=0; i<2; i++)
        for(int j=0; j<2; j++)
        {
            c.mapp[i][j]=0;
            for(int k=0; k<2; k++)
                c.mapp[i][j]+=(a.mapp[i][k]*b.mapp[k][j])%MOD;//mod[M];
            c.mapp[i][j]%=MOD;
        }
    return c;
}
int pow(int  n)
{
    base.mapp[0][0] =base.mapp[0][1]=base.mapp[1][0]=1;
    base.mapp[1][1]=0;
    ans.mapp[0][0] = ans.mapp[1][1] = 1;// ans 初始化为单位矩阵
    ans.mapp[0][1] = ans.mapp[1][0] = 0;
    while(n)
    {
        if(n&1)   ans=mult(ans,base);
        base=mult(base,base);
        n>>=1;
    }
    return ans.mapp[0][1];
}
int main()
{
    int t;
    while(cin>>N&&N!=-1)
    {
        printf("%d\n",pow(N));
    }
    return 0;
}