思路：

最後推出公式，發現只要用高斯消元求出矩陣A的逆即可

Code：

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
#include <array>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include 
 
<sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define
 
 slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 200 + 10;
const ll M = 4e12;
const int mod = 998244353;

int n;
ll a[N][N], b[N], p[N][N];
ll ans = 0;

ll qmi(ll a, ll b, ll p){
    ll res = 1;
    while(b){
        if(b & 1) res = res * a % p;
        a 
 
= a * a % p;
        b >>= 1;
    }
    return res;
}

void gauss()
{
    int c, r;
    for (c = 0, r = 0; c < n; c ++ )
    {
        int t = r;
        for (int i = r; i < n; i ++ ) 
            if (a[i][c] > 0){
                t = i;
                break;
            }

        if (a[t][c] == 0) return;

        for (int i = 0; i < n; i ++) {
            swap(a[t][i], a[r][i]);
            swap(p[t][i], p[r][i]);
        }       

        ll k = qmi(a[r][r], mod - 2, mod);
        for(int i = 0; i < n; i ++){
            a[r][i] = k * a[r][i] % mod;
            p[r][i] = k * p[r][i] % mod;
        }

        for (int i = 0; i < n; i ++){  
            if(i == r) continue;     
            if (a[i][c] != 0){
                k = a[i][c];
                for (int j = 0; j < n; j ++){
                    a[i][j] = (a[i][j] - a[r][j] * k % mod + mod) % mod;
                    p[i][j] = (p[i][j] - p[r][j] * k % mod + mod) % mod;;
                }
            }
        }
        r ++ ;
    }
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    freopen("/home/jungu/code/out.txt", "w", stdout);
    // freopen("/home/jungu/code/out.txt","w",stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    while(~sd(n)){  
        
        memset(p, 0, sizeof(p));

        for(int i = 0; i < n; i ++){
            p[i][i] = 1;
        }

        for(int i = 0; i < n; i ++){
            for(int j = 0; j < n; j ++){
                slld(a[i][j]);
                a[i][j] = (a[i][j] % mod + mod) % mod;
            }
        }

        for(int i = 0; i < n; i ++){
            slld(b[i]);
            b[i] = (b[i] % mod + mod) % mod;
        }
        
        gauss();

        ans = 0;
        for(int i = 0; i < n; i ++){
            for(int j = 0; j < n; j ++){
                ans = (ans + b[j] * p[j][i] % mod * b[i] % mod) % mod;
            }
        }
        cout << ans << endl;
        
    }

    return 0;
}