HDU 4549

来源：互联网发布：windows xp 下载编辑：程序博客网时间：2024/06/06 13:10

水题：费马小定理+快速幂+矩阵快速幂

（第一次用到费马小定理）

#include<bits/stdc++.h>using namespace std;typedef long long LL;const LL MOD  = 1000000006;const LL MOD1 = 1000000007;struct Matrix{    LL NUM[2][2];    Matrix operator + (const Matrix a) const    {        Matrix c;        for(int i = 0; i < 2; ++i)        {            for(int j = 0; j < 2; ++j)            {                c.NUM[i][j] = NUM[i][j] + a.NUM[i][j];            }        }        return c;    }    Matrix operator * (const Matrix a) const    {        Matrix c;        for(int i = 0; i < 2; ++i)        {            for(int j = 0; j < 2; ++j)            {                c.NUM[i][j] = 0;                for(int k = 0; k < 2; ++k)                    c.NUM[i][j] = (c.NUM[i][j] + NUM[i][k] * a.NUM[k][j] % MOD) % MOD;            }        }        return c;    }};Matrix ppow(Matrix a, LL n){    Matrix ret;    for(int i =0 ; i< 2; ++i)    {        for(int j = 0; j < 2; ++j)            ret.NUM[i][j] = i==j ? 1 : 0;    }    while(n)    {        if(n & 1) ret = ret * a;        a = a * a;        n >>= 1;    }    return ret;}LL Pow(LL a, LL n){    LL ret = 1;    while(n)    {        if(n & 1) ret =ret * a % MOD1;        a = a * a % MOD1;        n >>= 1;    }    return ret;}int main(){    LL a, b, n;    Matrix E;    E.NUM[0][0] = 1; E.NUM[0][1] = 1;    E.NUM[1][0] = 1; E.NUM[1][1] = 0;    while(cin >> a >> b >> n)    {        if(n == 0) cout << a << endl;        else if(n == 1) cout << b << endl;        else        {            n -= 1;            Matrix tmp = ppow(E,n);            LL na = tmp.NUM[0][1] , nb = tmp.NUM[0][0];            LL ans = (Pow(a,na) * Pow(b,nb))%MOD1;            cout << ans << endl;        }    }    return 0;}

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