矩阵快速幂求斐波那契

#include<bits/stdc++.h>
using namespace std;
const int p=10000;
typedef long long ll;
int n,m;

struct node {
    ll a[2][2];

    node operator*(const node &b) const {
        node res;
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                res.a[i][j] = 0;
                for (int k = 0; k < 2; k++) {
                    res.a[i][j] = (res.a[i][j] + a[i][k] * b.a[k][j]) % m;
                }
            }
        }
        return res;
    }
};

node pow(node b,ll c) {
    node res;
    res.a[0][0] = 1;
    res.a[1][0] = 0;
    res.a[0][1] = 0;
    res.a[1][1] = 1;
    while (c) {
        if (c & 1) {
            res = res * b;
        }
        c >>= 1;
        b = b * b;
    }
    return res;
}

int main() {
    node f;
    f.a[0][0] = 1;
    f.a[1][0] = 1;
    f.a[0][1] = 1;
    f.a[1][1] = 0;
    scanf("%d%d", &n, &m);
        node ans = pow(f, n);
        printf("%lld
", ans.a[1][0]);
}