CF450B【矩陣快速冪+負數取模】水題

寫給自己看的，記一下負數取模這個點

題目描述

f[1] = x, f[2] = y, f[i] = f[i-1] + f[i+1]，求第n項

思路

顯然f[i+1] = f[i] - f[i - 1] => f[i] = f[i - 1] - f[i - 2] 構造矩陣就好
注意負數的取模 ((num % mod) + mod) % mod

代碼

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

const int mod = 1e9 + 7;
const int N = 2;
struct Matrix {
    ll m[N][N];
    Matrix() {
        memset(m, 0, sizeof(m));
    }
};

Matrix mul(Matrix a, Matrix b)
{
    Matrix res;
    for(int i = 0; i < N; i++) {
        for(int j = 0; j < N; j++) {
            for(int k = 0; k < N; k++) {
                res.m[i][j] = (res.m[i][j] + a.m[i][k] * b.m[k][j]) % mod;
            }
        }
    }
    return res;
}

Matrix fastm(Matrix a, ll n)
{
    Matrix res;
    for(int i = 0; i < N; i++) {
        res.m[i][i] = 1;
    }
    while(n) {
        if(n & 1) res = mul(res, a);
        a = mul(a, a);
        n >>= 1;
    }
    return res;
}

void solve()
{
    ll x, y, n;
    cin >> x >> y >> n;
    Matrix a, b;
    if(n == 1) {
        cout << ((x % mod) + mod) % mod << endl;
        return;
    }
    if(n == 2) {
        cout << ((y % mod) + mod) % mod << endl;
        return;
    }
    a.m[0][0] = 1, a.m[0][1] = -1;
    a.m[1][0] = 1;
    a = fastm(a, n - 2);
    ll ans = (y * a.m[0][0] % mod + x * a.m[0][1] % mod) % mod;
    cout << ((ans % mod) + mod) % mod << endl;

}

int main()
{
    //freopen("in.txt", "r", stdin);
    solve();
    return 0;
}