2019牛客多校第五场B-generator 1(矩阵快速幂)

generator 1

题目传送门

解题思路

矩阵快速幂。只是平时的矩阵快速幂是二进制的，这题要用十进制的快速幂。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

inline int read(){
    int res = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -res : res;
}

ll mod;
struct Matrix{
    ll m[2][2];
    Matrix(){
        m[0][0] = m[0][1] = m[1][0] = m[1][1] = 0;
    }
    Matrix operator*(const Matrix& a)const{
        Matrix ans;
        for(int i = 0; i < 2; i ++){
            for(int j = 0; j < 2; j ++)
                for(int k = 0; k < 2; k ++)
                    ans.m[i][j] = (ans.m[i][j] + m[i][k] * a.m[k][j]) % mod;
        }
        return ans;
    }
};

Matrix fpow(const Matrix& x, string str)
{
    int k = str.size() - 1;
    Matrix t = x;
    Matrix ans;
    ans.m[0][0] = 1, ans.m[1][1] = 1;
    while(k >= 0){
        for(int i = 1; i <= str[k] - '0'; i ++)
            ans = ans * t;
        Matrix temp = t * t;
        t = temp * temp;
        t = t * t * temp;
        --k;
    }
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    ll x0, x1, a, b;
    string n;
    cin >> x0 >> x1 >> a >> b >> n >> mod;
    Matrix t;
    t.m[0][0] = a, t.m[0][1] = b, t.m[1][0] = 1;
    Matrix ans = fpow(t, n);
    printf("%lld
", (ans.m[1][0] * x1 + ans.m[1][1] * x0) % mod);
    return 0;
}