牛客 110D 矩阵

假设$C=AB$, 那么答案就为

$\begin{align} \notag ans & =\sum\limits_{i=0}^{n-1}\sum\limits_{j=0}^{n-1}C[i][j]p^{(n-i)n-1-j} \\ \notag & =  \sum\limits_{i=0}^{n-1}\sum\limits_{j=0}^{n-1}\sum\limits_{k=0}^{n-1}A[i][k]B[k][j]p^{(n-i)n-1-j} \\ & = \sum\limits_{k=0}^{n-1}\Big(\sum\limits_{i=0}^{n-1}A[i][k]p^{(n-i)n-1}\Big)\Big(\sum\limits_{j=0}^{n-1}B[k][j]p^{-j}\Big) \notag \end{align}$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

int n, Aa, Ab, Ac, Ad, Ba, Bb, Bc, Bd, p;
uint32_t x, y, z, w;
uint32_t xorshift() {
    uint32_t t = x;
    t ^= t << 11;
    t ^= t >> 8;
    x = y; y = z; z = w;
    w ^= w >> 19;
    w ^= t;
    return w & ((1 << 24) - 1);
}
void get(uint32_t a, uint32_t b, uint32_t c, uint32_t d) {
    x = a; y = b; z = c; w = d;
}

const int N = 7e3+10;
int f1[N], f2[N], g1[N], g2[N];
int main() {
	scanf("%d%d%d%d%d%d%d%d%d%d", &n, &Aa, &Ab, &Ac, &Ad, &Ba, &Bb, &Bc, &Bd, &p);
	int w1 = inv(qpow(p,n)), w2 = inv(p);
	int r1 = qpow(p,n*n-1), r2 = 1;
	REP(i,0,n-1) {
		f1[i]=r1,r1=(ll)r1*w1%P;
		f2[i]=r2,r2=(ll)r2*w2%P;
	}
	get(Aa,Ab,Ac,Ad);
	REP(i,0,n-1) REP(j,0,n-1) {
		g1[j] = (g1[j]+(ll)xorshift()*f1[i])%P;
	}
	get(Ba,Bb,Bc,Bd);
	REP(i,0,n-1) REP(j,0,n-1) {
		g2[i] = (g2[i]+(ll)xorshift()*f2[j])%P;
	}
	int ans = 0;
	REP(i,0,n-1) ans = (ans+(ll)g1[i]*g2[i])%P;
	if (ans<0) ans += P;
	printf("%d\n", ans);
}