题解 | #字符框#

思路

遍历字符数组，考虑以当前遍历字符为左上元素的2*2矩阵是否符合规则即可

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

int n,m,cot,res = 0;
char car[52][52];
bool jgface(int x,int y){    //利用质因数分解的唯一性
    cot = 1;
    for(int i = x;i <= x+1;i++)
        for(int j = y;j <= y+1;j++){
            if(car[i][j] == 'f') cot *= 2;
            if(car[i][j] == 'a') cot *= 3;
            if(car[i][j] == 'c') cot *= 5;
            if(car[i][j] == 'e') cot *= 7;
        }
    if(cot == 210) return true;
    else return false;
}

int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    cin>>n>>m;
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= m;j++)
            cin>>car[i][j];
    
    for(int i = 1;i <= n-1;i++)    //判断以i,j为左上顶点的2*2矩形是否满足要求
        for(int j = 1;j <= m-1;j++)    //遍历至边界-1即停，省去边界判断
            if(jgface(i,j)) res++;
    cout<<res;
    return 0;
}