---
title: "L1-054 福到了"
created: 2025-11-28
tags:
  - 算法
---

# L1-054 福到了

## 题目 [L1-054 福到了](https://pintia.cn/problem-sets/994805046380707840/exam/problems/type/7?problemSetProblemId=994805076512587776&page=0)

![[image-a8c8fd56.png]]

## 思路分析

## 代码实现

```cpp
#include<bits/stdc++.h>

using namespace std;

#define endl '\n'

using ll = long long;

using ull = unsigned long long;

using PII = pair<int,int>;

using Pll = pair<ll,ll>;

int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};

const int inf = 0x3f3f3f3f;

int main(){

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

	char c;int n;cin>>c>>n;

	cin.ignore();

	vector<vector<char>> g(n);

	for(int i=0;i<n;i++){

		string s; getline(cin,s);

		for(int j=0;j<n;j++){

			g[i].push_back(s[j]);

		}

	}

	bool buyongdao=true;

	for(int i=0;i<n;i++){

		for(int j=0;j<n;j++){

//			if(g[i][j]==' ')	cout<<g[i][j];

//			else if(g[i][j]=='@')	cout<<c;

			if(g[i][j]!=g[n-i-1][n-j-1]){

				buyongdao=false;

			}

		}

	}

	if(buyongdao){

		cout<<"bu yong dao le"<<endl;

	}

	for(int i=0;i<n;i++){

		for(int j=0;j<n;j++){

			if(g[n-i-1][n-j-1]=='@')	cout<<c;

			else cout<<g[n-i-1][n-j-1];

		}

		cout<<endl;

	}

	return 0;

}
```

## 同类题型

## 视频讲解

---
⬅️ [[L1-053 电子汪|L1-053 电子汪]] 🏠 [[00-天梯赛]] ➡️ [[L1-055 谁是赢家|L1-055 谁是赢家]]