HDU 4565矩阵快速幂—— So Easy!

来源：互联网发布：艾瑞咨询数据编辑：程序博客网时间：2024/06/05 08:23

题目链接

借鉴别人的一张解题思路
这里写图片描述

转化成了 (a^n + b^n) %M

#include<iostream>#include<string>#include<vector>#include<algorithm>#include<queue>#include<cstdio>#include<cstring>#include<cmath>#include<map>using namespace std;typedef  long long ll;ll MOD ;struct Matrix {    ll a[2][2];    Matrix() {        memset(a, 0, sizeof(a));    }    Matrix operator * (const Matrix y) {        Matrix ans;        for(int i = 0; i <= 1; i++)            for(int j = 0; j <= 1; j++)                for(int k = 0; k <= 1; k++)                    ans.a[i][j] += a[i][k]*y.a[k][j];        for(int i = 0; i <= 1; i++)            for(int j = 0; j <= 1; j++)                ans.a[i][j] %= MOD;        return ans;    }    void operator = (const Matrix b) {        for(int i = 0; i <= 1; i++)            for(int j = 0; j <= 1; j++)                a[i][j] = b.a[i][j];    }};ll solve(ll a,ll b,ll n) {    Matrix ans, trs;    ans.a[0][1] = 1;    //初始值    ans.a[0][0] = a;    ans.a[0][1] = 2;    // f(n) = a*f(n-1) + b * f(n-2);    trs.a[0][0] = a;    trs.a[1][0] = -b;    trs.a[0][1] = 1;    while(n) {        if(n&1)            ans = ans*trs;        trs = trs*trs;        n >>= 1;    }    return (ans.a[0][1]+ans.a[1][1]+MOD)%MOD;//improtant}int main() {     ios_base::sync_with_stdio(false);//    freopen("data.txt","r",stdin);    ll aa,bb,m,n;    while(cin>>aa>>bb>>n>>m)    {        ll p = 2*aa;        ll q = aa*aa-bb;        MOD  = m;        cout<<solve(p,q,n)<<endl;    }    return 0;}

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