矩阵快速幂 斐波那契数列

离散数论知识点：

（a+b）%c = （a%c）+(b%c)

（a*b）%c  = (（a%c）*（b%c）)%c

快速幂：

int quickpow(int m,int n,int k){    int b = 1;    while(n>0){        if(n & 1)            b = (b*m)%k;        n = >>1;        m = (m*m)%k;    }    return b;}

求斐波拉契数列的某项的前四位和后四位：

#include <iostream>#include <cstdio>#include <cmath>#define Mod 10000using namespace std;const int MAX=2;long long f[100];typedef struct {    int m[MAX][MAX];}MATRIX;MATRIX P = {0,1,1,1};MATRIX I = {1,0,0,1};//矩阵相乘MATRIX matrixmul(MATRIX a,MATRIX b){    int i,j,k;    MATRIX c;    for(i=0;i<MAX;i++){        for(j=0;j<MAX;j++){            c.m[i][j] = 0;            for(k=0;k<MAX;k++){                c.m[i][j] += (a.m[i][k]*b.m[k][j])%Mod;            }            c.m[i][j] %= Mod;        }    }    return c;}//矩阵的快速幂MATRIX quickpow(long long n){    MATRIX b = I;    MATRIX m = P;    while(n>0){        if(n&1)            b = matrixmul(b,m);        n = n>>1;        m = matrixmul(m,m);    }    return b;}int main(){    double log_s = 0.0;    MATRIX temp;    int n,i,bit=0;    f[0]=0;    f[1]=1;    for(i=2;i<=50;i++){        f[i] = f[i-1]+f[i-2];        if(f[i]>100000000){            bit=i-1;            break;        }    }    while(cin>>n){        if(n<=bit)            cout<<f[n]<<endl;        if(n>bit){            temp = quickpow(n);            int last_bit = temp.m[0][1];            log_s=log10(1.0/sqrt(5)) +(double)n*log10((1.0+sqrt(5))/2.0);            int ans_s=(int)(pow(10,log_s-(int)log_s+3));            cout<<ans_s<<"...";            printf("%04d\n",last_bit);        }    }    return 0;}