虽然非常牛波一的申哥已经发了一次题解，但是他的代码都是挤在main函数里，可能导致有的同学理解起来有些困难，特此水一发自己的

Ps：因为没看清题卡了好久

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

typedef long long ll;
const ll N = 3e2 + 10;

ll n, m, num[N][N], dui[N][N], tmp[N][N], res[N][N];
char code[N][N];
string ans;

bool isPrime(ll x){             // 判断是否为质数
    if (x == 2 || x == 3 || x == 5 || x == 7) return true;
    return false;
}

bool isUpper(char x){           // 判断是否大写
    return (x >= 'A' && x <= 'Z');
}

bool isCorre(ll x, ll y){       // 判断该区域是否有效，x和 y是字母表中要枚举的矩阵的开始下标
    ll x2 = x + n - 1, y2 = y + n - 1;

    if (num[1][1] == dui[x][y] && num[1][n] == dui[x][y2] && num[n][1] == dui[x2][y] && num[n][n] == dui[x2][y2]) return true;
    return false;
}

void turnRight(ll cnt){         // 把数字表旋转，cnt是要转几次
    while (cnt --){
        for (ll i = 1; i <= n; i ++)
            for (ll j = 1; j <= n; j ++)
                tmp[i][j] = num[n - j + 1][i];
        
        for (ll i = 1; i <= n; i ++)
            for (ll j = 1; j <= n; j ++)
                num[i][j] = tmp[i][j];
    }
}

ll countUpper(ll x, ll y){      // 统计 4个角的大写字母数量
    ll x2 = x + n - 1, y2 = y + n - 1;

    return ((isUpper(code[x][y])) + (isUpper(code[x][y2])) + (isUpper(code[x2][y])) + (isUpper(code[x2][y2])));
}

void decode(ll x, ll y){       // 解密，转换成密码
    string tmpans;

    for (ll i = x; i <= x + n - 1; i ++){
        for (ll j = y; j <= y + n - 1; j ++){
            if ((isPrime(num[i - x + 1][j - y + 1]) == isUpper(code[i][j])) && num[i - x + 1][j - y + 1] != 1 || (!num[i - x + 1][j - y + 1])) tmpans += code[i][j];
            else continue;
        }
    }

    ans += tmpans;
}

void restore(){                 // 转完后，把数字表还原
    for (ll i = 1; i <= n; i ++)
        for (ll j = 1; j <= n; j ++) num[i][j] = res[i][j];

    return;
}

int main(){
    ios::sync_with_stdio(0), cout.tie(0), cin.tie(0);

    cin >> n >> m;

    // 输入数字表
    for (ll i = 1; i <= n; i ++) 
        for (ll j = 1; j <= n; j ++){
            char x;
            cin >> x;
            num[i][j] = x - '0', res[i][j] = num[i][j];
        }

    // 输入字母表
    for (ll i = 1; i <= m; i ++) 
        for (ll j = 1; j <= m; j ++){
            cin >> code[i][j];
            if (code[i][j] >= 'A' && code[i][j] <= 'Z') dui[i][j] = code[i][j] - 'A' + 1;
            else dui[i][j] = code[i][j] - 'a' + 1;
        }

    // 枚举有效区域
    for (ll i = 1; i <= m - n + 1; i ++)    
        for (ll j = 1; j <= m - n + 1; j ++){
            if (isCorre(i, j)){
                ll uppercnt = countUpper(i, j);
                do{         // 根据题目要求解密
                    decode(i, j);
                    turnRight(1);
                }while (uppercnt --);

                restore();  // 解密完后把数字表回去
            }
        }


    // 输出答案
    if (!ans.size()) cout << "No solution";
    else cout << ans << endl;

    return 0;
}

结合注释可以更好地理解一下，$Goodbye.$