矩阵快速幂

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复杂度为o（n^3logk）

/*求 a^k % mod，其中a是n*n的矩阵*/const int mod = 10000;const int maxn = 2;_LL k;int n;struct matrix{    _LL mat[maxn][maxn];} a,res;matrix mul(matrix x, matrix y){    matrix tmp;    memset(tmp.mat,0,sizeof(tmp.mat));    for(int i = 0; i < n; i++)    {        for(int k = 0; k < n; k++)        {            if(x.mat[i][k] == 0) continue; //小小的优化。            for(int j = 0; j < n; j++)            {                tmp.mat[i][j] += x.mat[i][k] * y.mat[k][j];                if(tmp.mat[i][j] >= mod)                    tmp.mat[i][j] %= mod;            }        }    }    return tmp;}void solve(){for(int i = 0; i < n; i++)for(int j = 0; j < n; j++)res.mat[i][j] = (i == j); //初始化为单位矩阵    while(k)    {        if(k & 1)            res = mul(res,a);        a = mul(a,a);        k >>= 1;    }}

http://poj.org/problem?id=3070

题意：当n非常大时，求斐波那契数列的第n项。矩阵快速幂模板。

#include <stdio.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#define LL long long#define _LL __int64#define eps 1e-8using namespace std;const int mod = 10000;struct matrix{    _LL mat[2][2];}a,res;_LL k;int n;matrix mul(matrix x, matrix y){    matrix tmp;    memset(tmp.mat,0,sizeof(tmp.mat));    for(int i = 0; i < n; i++)    {        for(int k = 0; k < n; k++)        {            if(x.mat[i][k] == 0) continue;            for(int j = 0; j < n; j++)            {                tmp.mat[i][j] += x.mat[i][k] * y.mat[k][j];                if(tmp.mat[i][j] >= mod)                    tmp.mat[i][j] %= mod;            }        }    }    return tmp;}int main(){    while(~scanf("%I64d",&k))    {        if(k == -1) break;        a.mat[0][0] = a.mat[0][1] = a.mat[1][0] = 1;        a.mat[1][1] = 0;        n = 2;        for(int i = 0; i < n; i++)        {            for(int j = 0; j < n; j++)            {                if(i == j) res.mat[i][j] = 1;                else res.mat[i][j] = 0;            }        }        while(k)        {            if(k & 1)                res = mul(res,a);            a = mul(a,a);            k >>= 1;        }        int ans = res.mat[0][1] % mod;        printf("%d\n",ans);    }    return 0;}

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