poj 3070 Fibonacci

题目链接：Fibonacci

题目大意：求斐波那契的第n项对10000求余

题目思路：构造矩阵后直接矩阵快速幂

#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const ll maxn = 3;const ll MOD = 10000;#define mod(x) ((x)%MOD)struct mat{    ll m[maxn][maxn];}unit;mat operator *(mat a,mat b){    mat ret;    ll x;    for(ll i = 0;i < maxn;i++){        for(ll j = 0;j < maxn;j++){            x = 0;            for(ll k = 0;k < maxn;k++)                x += mod((ll)a.m[i][k]*b.m[k][j]);            ret.m[i][j] = mod(x);        }    }    return ret;}void init_unit(){    for(ll i = 0;i < maxn;i++)        unit.m[i][i] = 1;    return ;}mat pow_mat(mat a,ll n){    mat ret = unit;    while(n){        if(n&1) ret = ret*a;        a = a*a;        n >>= 1;    }    return ret;}int main(){    int n;    init_unit();    while(~scanf("%d",&n)&&n!=-1){        if(n == 0) puts("0");        else if(n == 1) puts("1");        else if(n == 2) puts("1");        else{            mat a,b;            b.m[0][0] = 1,b.m[0][1] = 1,b.m[0][2] = 0;            b.m[1][0] = 1,b.m[1][1] = 0,b.m[1][2] = 1;            b.m[2][0] = 0,b.m[2][1] = 0,b.m[2][2] = 0;            a.m[0][0] = 1,a.m[0][1] = 1,a.m[0][2] = 0;            b = pow_mat(b,n-2);            a = a*b;            printf("%lld\n",a.m[0][0]%MOD);        }    }    return 0;}