[BZOJ 4128]Matrix

来源：互联网发布：大数据运维管理平台编辑：程序博客网时间：2024/06/06 11:22
裸BSGS矩阵求逆。。
#include <bits/stdc++.h>using namespace std;typedef long long ll;int n, md;#define maxn 72#define mod 13331struct Matrix{    int a[maxn][maxn];    void read(){        for(int i = 0; i < n; i ++)            for(int j = 0; j < n; j ++)                scanf("%d", &a[i][j]);    }    void print(){        for(int i = 0; i < n; i ++){            for(int j = 0; j < n; j ++)                printf("%d ", a[i][j]);            printf("\n");        }    }    unsigned long long hash(){        unsigned long long ret = 0;        for(int i = 0; i < n; i ++)            for(int j = 0; j < n; j ++)                ret = ret * mod + a[i][j];        return ret;    }    void clear(){memset(a, 0, sizeof a);}    void set(){clear(); for(int i = 0; i < n; i ++)a[i][i] = 1;}}A, B, mat;Matrix operator*(const Matrix& a, const Matrix& b){    Matrix c; c.clear();    for(int i = 0; i < n; i ++)        for(int j = 0; j < n; j ++)            for(int k = 0; k < n; k ++)                c.a[i][j] = (c.a[i][j] + (long long)a.a[i][k] * b.a[k][j]) % md;    return c;}map<unsigned long long, int> M;ll power_mod(ll a, ll b){    ll ret = 1;    while(b > 0){        if(b & 1)ret = ret * a % md;        b >>= 1;        a = a * a % md;    }return ret;}Matrix getinv(Matrix a){    Matrix ret; ret.set();    for(int i = 0; i < n; i ++){        for(int j = i; j < n; j ++){            if(a.a[j][i]){                for(int k = 0; k < n; k ++){                    swap(a.a[i][k], a.a[j][k]);                    swap(ret.a[i][k], ret.a[j][k]);                }                ll inv = power_mod(a.a[i][i], md-2);                for(int k = 0; k < n; k ++){                    a.a[i][k] = a.a[i][k] * inv % md;                    ret.a[i][k] = ret.a[i][k] * inv % md;                }                break;            }        }        //if(!a.a[i][i])continue;        for(int j = 0; j < n; j ++){            if(i == j)continue;            ll tmp = a.a[j][i];            for(int k = 0; k < n; k ++){                a.a[j][k] = (a.a[j][k] - a.a[i][k] * tmp % md + md) % md;                 ret.a[j][k] = (ret.a[j][k] - ret.a[i][k] * tmp % md + md) % md;            }        }    }    return ret;}int main(){    scanf("%d%d", &n, &md);    A.read(), B.read();    int q = sqrt(md) + 1;     mat.set();    unsigned long long now = B.hash(), tmp = mat.hash();    if(tmp == now)return puts("0"), 0;    for(int i = 1; i <= q; i ++){        mat = mat * A;        tmp = mat.hash();        if(tmp == now){            printf("%d\n", i);            return 0;        }        M[tmp] = i;    }    Matrix inv = getinv(mat);    for(int i = 1; i <= q; i ++){        B = B * inv;        now = B.hash();        if(M.count(now)){            printf("%d\n", i * q + M[now]);            return 0;        }    }    return 0;}
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