Character Wheels

鏈接：https://ac.nowcoder.com/acm/contest/8688/I
來源：牛客網

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 524288K，其他語言1048576K
64bit IO Format: %lld

題目描述

You are given one n \times nn×n character wheels, and it is guaranteed that n is even. Please implement these following operations:

Clockwise/Counter-clockwise rotate the x-th wheel y(1≤y≤109 ) times. Rotating one time means rotatting 90 degrees. The instruction is L/R x y, R indicates rotating clockwise and L indicates rotating counter-clockwise. For example, R 1 3 means rotate the 1-st wheel 3 times clockwise.
Print all the wheels, the instruction is P.

輸入描述

First line contains one integer 𝑛 ( 4 ≤ 𝑛 ≤ 50), indicates the size of the wheels.
Then followed the wheels.
After that, one line contains one integer 𝑚 ( 1 ≤ 𝑚 ≤ 100), indicates the number ofoperations.
Then followed 𝑚 lines, one line one instruction.
it is guaranteed that at least there is one P instruction and the wheels only contain lowercase letters.

輸出描述

If the instruction is P, then print all the wheels.

題目大意

給一個n*n的矩陣，從外到內每一圈編號1~n/2，輸入R就是向右轉，L就是向左轉，將第x圈旋轉y次，每次旋轉90°，注意題中要求y的次數小于10的九次方，因此需要y % 4 簡化次數，避免超時
輸入P列印當前矩陣

示例1

輸入

4
abcd
cabe
fgha
edbe 
5
L 1 1 
P
R 1 3
L 2 1 
P

輸出

deae
cabb
bghd
acfe
ebde
abhf
eagc
dcba

思路

先把矩陣記錄下來，
定義一個二位陣列存入每個環的字符順序，（均從一號位開始存入）a[環層數][0] 初始為0
之后根據具體旋轉情況，改變a[環層數][0] 的 值，R1向右轉加一，L1向左轉減一，
輸入P就將二維陣列從新回傳矩陣，將矩陣輸出即可，

C++ 代碼

#include<iostream>
#include<cstring>
using namespace std;

#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

const int N = 10005;
char ma[55][55];
char a[55][N];
int n,T;

void print(){
	for(int i = 1; i <= n/2; i ++)
	{
		int m = 1;
		if(a[i][0] == '1'){
			int hang = i,lie = i;
			do{ma[hang][lie] = a[i][m++]; lie ++;}while(m <= n - 2 * i + 1);
			do{ma[hang][lie] = a[i][m++]; hang++;}while(m <= 2*n - 4*i + 2);
			do{ma[hang][lie] = a[i][m++]; lie--;}while(m <= 3*n - 6*i + 3);
			do{ma[hang][lie] = a[i][m++]; hang--; }while(m <= 4*n - 8*i + 4);
		}
		else if(a[i][0] == '2'){
			int hang = i,lie = n - i + 1;
			do{ma[hang][lie] = a[i][m++]; hang ++;}while(m <= n - 2 * i + 1);
			do{ma[hang][lie] = a[i][m++]; lie --;}while(m <= 2*n - 4*i + 2);
			do{ma[hang][lie] = a[i][m++]; hang --;}while(m <= 3*n - 6*i + 3);
			do{ma[hang][lie] = a[i][m++]; lie ++; }while(m <= 4*n - 8*i + 4);
		}
		else if(a[i][0] == '3'){
			int hang = n - i + 1,lie = n - i + 1;
			do{ma[hang][lie] = a[i][m++]; lie --;}while(m <= n - 2 * i + 1);
			do{ma[hang][lie] = a[i][m++]; hang --;}while(m <= 2*n - 4*i + 2);
			do{ma[hang][lie] = a[i][m++]; lie ++;}while(m <= 3*n - 6*i + 3);
			do{ma[hang][lie] = a[i][m++]; hang ++; }while(m <= 4*n - 8*i + 4);
		}
		else if(a[i][0] == '4'){
			int hang = n - i + 1,lie = i;
			do{ma[hang][lie] = a[i][m++]; hang --;}while(m <= n - 2 * i + 1);
			do{ma[hang][lie] = a[i][m++]; lie ++;}while(m <= 2*n - 4*i + 2);
			do{ma[hang][lie] = a[i][m++]; hang ++;}while(m <= 3*n - 6*i + 3);
			do{ma[hang][lie] = a[i][m++]; lie --; }while(m <= 4*n - 8*i + 4);
		}
	}
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= n; j ++)
		{
			cout << ma[i][j];
		}
		cout << endl;
	}
}

int main()
{
	IO;
	memset(a,0,sizeof(a));
	cin >> n;
	for(int i = 1; i <= n; i ++)
	for(int j = 1; j <= n; j ++)
		cin >> ma[i][j];
	for(int i = 1; i <= n/2; i ++)
	{
		a[i][0] = '1';
		int m = 1,hang = i,lie = i;
		do{a[i][m++] = ma[hang][lie]; lie++;}while(m <= n - 2 * i + 1);//m <= (n - 2*(i - 1) - 1)   //向右 
		do{a[i][m++] = ma[hang][lie]; hang++;}while(m <= 2*n - 4*i + 2);//m <= 2*(n - 2*(i - 1) - 1)  //向下 
		do{a[i][m++] = ma[hang][lie]; lie--;}while(m <= 3*n - 6*i + 3);//m <= 3*(n - 2*(i - 1) - 1)  //向左 
		do{a[i][m++] = ma[hang][lie]; hang--; }while(m <= 4*n - 8*i + 4);//m <= 4*(n - 2*(i - 1) - 1) //向上 
	}
	

	cin >> T;
	while(T --)
	{
		char ch;
		int x,y;
		cin >> ch;
		getchar();
		if(ch == 'L'){
			cin >> x >> y;
			y = y % 4;
			while(y--){
				a[x][0] --;
				if(a[x][0] < '1')a[x][0] += 4; 
			}
		}
		else if(ch == 'R'){
			cin >> x >> y;
			y = y % 4;
			while(y--){
				a[x][0] ++;
				if(a[x][0] > '4')a[x][0] -= 4;
			}
		}
		else if(ch == 'P'){
			print();
		}
	}	
	
//	for(int j = 1; j <= n/2; j ++)
//	{
//		for(int i = 1; i <= 4*(n-1); i ++)
//			cout << a[j][i] << " ";
//		cout << endl;
//	}

	
	
	return 0;
}