[编程题]C翻转
- 热度指数：5714 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 64M，其他语言128M
- 算法知识视频讲解
首先输入一个5 * 5的数组，然后输入一行，这一行有四个数，前两个代表操作类型，后两个数x y代表需操作数据为以x y为左上角的那几个数据。 操作类型有四种： 1 2 表示：90度，顺时针，翻转4个数 1 3 表示：90度，顺时针，翻转9个数 2 2 表示：90度，逆时针，翻转4个数 2 3 表示：90度，逆时针，翻转9个数
输入描述:
输入有多组数据。 每组输入一个5 * 5的数组，然后输入一行，这一行有四个数，前两个代表操作类型，后两个数x y代表需操作数据为以x y为左上角的那几个数据。
输出描述:
输出翻转后的数组。
示例1
输入
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 1 3 1 1
输出
11 6 1 4 5 12 7 2 9 10 13 8 3 14 15 16 17 18 19 20 21 22 23 24 25
将矩阵旋转的通用方法：
1.顺时针旋转：先将矩阵上下翻转，再将矩阵元素沿对角线对称交换
2.逆时针旋转：先将矩阵左右翻转，再将矩阵元素沿对角线对称交换
#include <iostream>
#include <algorithm>
using namespace std;
// 上下翻转
void flipTop2Down(int(&mat)[5][5], int x, int y, int len) {
int down = x + len - 1;
while (down > x) {
for (int i = y; i < y + len; ++i)
swap(mat[x][i], mat[down][i]);
++x, --down;
}
}
// 左右翻转
void flipLeft2Right(int(&mat)[5][5], int x, int y, int len) {
int right = y + len - 1;
while (y < right) {
for (int i = x; i < x + len; ++i)
swap(mat[i][y], mat[i][right]);
++y, --right;
}
}
// 对称
void sym(int(&mat)[5][5], int x, int y, int len) {
for (int i = 0; i < len; ++i) {
for (int j = i + 1; j < len; ++j) {
swap(mat[x+i][y+j], mat[x+j][y+i]);
}
}
}
void print(int(&mat)[5][5]) {
for (int i = 0; i < 5; ++i) {
for (int j = 0; j < 5; ++j) {
cout << mat[i][j] << ' ';
}
cout << endl;
}
}
int main() {
int mat[5][5];
int direct, len, x, y;
for (int i = 0; i < 5; ++i) {
for (int j = 0; j < 5; ++j) {
cin >> mat[i][j];
}
}
cin >> direct >> len >> x >> y;
x = x - 1;
y = y - 1;
// 顺时针旋转
if (direct == 1) {
// 上下翻转
flipTop2Down(mat, x, y, len);
// 对称交换
sym(mat, x, y, len);
}
else {
// 左右翻转
flipLeft2Right(mat, x, y, len);
// 对称交换
sym(mat, x, y, len);
}
print(mat);
}
编辑于 2019-03-19 21:55:21
回复(1)
package com.speical.first;
import java.util.Scanner;
/**
* C翻转
*
* 很简单，模拟正常翻转就可以了
*
* 我们可以发现当顺时针90的时候
* 第一行的数据依次为第一列的倒序
* 第二行的数据依次为第二列的倒序
* ....
*
* 我们可以发现当逆时针90的时候
* 第一行的数据依次为最后一列的正序
* 第二行的数据依次为倒数第二列的正序
* ....
*
* So, it it easy that you can handle it!
* @author special
* @date 2018年2月3日 下午4:18:16
*/
public class Pro181 {
static int[][] nums = new int[5][5];
static int[][] auxiliary = new int[5][5];
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
while(input.hasNext()){
for(int i = 0; i < 5; i++){
for(int j = 0; j < 5; j++){
nums[i][j] = input.nextInt();
auxiliary[i][j] = nums[i][j];
}
}
int kind = input.nextInt();
int count = input.nextInt();
int startX = input.nextInt() - 1; //数组的索引从0开始，所以此处减一
int startY = input.nextInt() - 1;
int endX = startX + count - 1;
int endY = startY + count - 1;
//旋转
for(int i = startX; i <= endX; i++){
for(int j = startY; j <= endY; j++){
if(kind == 1){
nums[i][j] = auxiliary[endX - (j - startY)][startY + (i - startX)];
}else if(kind == 2){
nums[i][j] = auxiliary[startX + (j - startY)][endY - (i - startX)];
}
}
}
//输出数组
for(int i = 0; i < 5; i++){
for(int j = 0; j < 5; j++){
System.out.print((j == 0 ? "" : " ") + nums[i][j]);
}
System.out.println();
}
}
}
}
发表于 2018-02-03 17:24:07
回复(4)
#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;
int main()
{
string str1, str2;
cin >> str1;
getchar(); //这一步很重要
while (getline(cin,str2))
{
int i = 0, j = 0;
for (;i+j < str2.size();)
{
if ((str2[i + j]|32) == (str1[j]|32))
j++;
else
{
i++;
j = 0;
}
if (j == str1.size())
{
for (int z = i;z < i + str1.size();++z)
str2[z] = ' ';
i = i + str1.size();
j = 0;
}
}
for (int i = 0;i < str2.size();++i)
if (str2[i] == ' ')
continue;
else
cout << str2[i];
cout << endl;
}
return 0;
}
编辑于 2017-02-26 11:23:32
回复(1)
#include <iostream>
#include <algorithm>
using namespace std;
void ri90(int tmp[5][5], int flag, int x, int y)
{
if(flag==2)
{
swap(tmp[x][y],tmp[x][y+1]);
swap(tmp[x+1][y],tmp[x+1][y+1]);
swap(tmp[x][y],tmp[x+1][y+1]);
}
else
{
swap(tmp[x][y],tmp[x][y+2]);
swap(tmp[x+2][y],tmp[x+2][y+2]);
swap(tmp[x][y],tmp[x+2][y+2]);
swap(tmp[x][y+1],tmp[x+1][y+2]);
swap(tmp[x+1][y],tmp[x+2][y+1]);
swap(tmp[x][y+1],tmp[x+2][y+1]);
}
}
void le90(int tmp[5][5], int flag, int x, int y)
{
if(flag==2)
{
swap(tmp[x][y],tmp[x][y+1]);
swap(tmp[x+1][y],tmp[x+1][y+1]);
swap(tmp[x][y+1],tmp[x+1][y]);
}
else
{
swap(tmp[x][y],tmp[x][y+2]);
swap(tmp[x+2][y],tmp[x+2][y+2]);
swap(tmp[x][y+2],tmp[x+2][y]);
swap(tmp[x][y+1],tmp[x+1][y+2]);
swap(tmp[x+1][y],tmp[x+2][y+1]);
swap(tmp[x+1][y],tmp[x+1][y+2]);
}
}
int main()
{
int tmp[5][5],i,j;
for(i=0;i<5;++i)
for(j=0;j<5;++j)
cin >> tmp[i][j];
int m,n,x,y;
cin>>m>>n>>x>>y;
--x;
--y;
if(m==1)
ri90(tmp,n,x,y);
else le90(tmp,n,x,y);
for(i=0;i<5;++i)
for(j=0;j<5;++j)
{
cout<<tmp[i][j];
if(j==4)
cout<<endl;
else cout<<" ";
}
return 0;
}
发表于 2019-04-27 12:31:43
回复(0)
#include <stdio.h>
#include <stdlib.h>
// m 控制旋转方向 顺时针/逆时针
// n 控制旋转的数据个数，n表示需要旋转的方阵行数或列数(n*n即表示需要旋转的数据个数)
// 题目中(x-1,y-1)表示需要旋转的二维数组最左上角元素的坐标
void reverse(int (*a)[5],int m,int n,int row,int col);
int main()
{
int i, j, m, n, row, col;
int a[5][5];
while(scanf("%d", &a[0][0]) != EOF)
{
//接收矩阵
for(i = 0; i < 5; i ++)
for(j = 0; j < 5; j ++)
{
if(i == 0 && j == 0)
continue;
else
scanf("%d", &a[i][j]);
}
//接收旋转参数
scanf("%d%d%d%d", &m, &n, &row, &col);
//由于数组是从0开始存储
row -= 1;
col -= 1;
//矩阵翻转
reverse(a,m, n, row, col);
//打印输出
for(i = 0; i < 5; i ++)
for(j = 0; j < 5; j ++)
{
if(j == 4)
printf("%d\n", a[i][j]);
else
printf("%d ", a[i][j]);
}
}
return 0;
}
// 翻转函数
void reverse(int (*a)[5],int m,int n,int row,int col)
{
int i,j;
int temp[n][n];
//判断旋转方向
for(i = 0; i < n; ++i)
for (j = 0; j < n;++j)
{
if(m == 1) //顺时针旋转90°
temp[j][n - i-1] = a[row + i][col + j];
else //逆时针旋转90°
temp[n-j-1][i] = a[row + i][col + j];
}
//替换a中对应旋转位置的数据
for(i=row;i<(row+n);i++)
for(j=col;j<(col+n);j++)
a[i][j]=temp[i-row][j-col];
}
#include <stdlib.h>
// m 控制旋转方向 顺时针/逆时针
// n 控制旋转的数据个数，n表示需要旋转的方阵行数或列数(n*n即表示需要旋转的数据个数)
// 题目中(x-1,y-1)表示需要旋转的二维数组最左上角元素的坐标
void reverse(int (*a)[5],int m,int n,int row,int col);
int main()
{
int i, j, m, n, row, col;
int a[5][5];
while(scanf("%d", &a[0][0]) != EOF)
{
//接收矩阵
for(i = 0; i < 5; i ++)
for(j = 0; j < 5; j ++)
{
if(i == 0 && j == 0)
continue;
else
scanf("%d", &a[i][j]);
}
//接收旋转参数
scanf("%d%d%d%d", &m, &n, &row, &col);
//由于数组是从0开始存储
row -= 1;
col -= 1;
//矩阵翻转
reverse(a,m, n, row, col);
//打印输出
for(i = 0; i < 5; i ++)
for(j = 0; j < 5; j ++)
{
if(j == 4)
printf("%d\n", a[i][j]);
else
printf("%d ", a[i][j]);
}
}
return 0;
}
// 翻转函数
void reverse(int (*a)[5],int m,int n,int row,int col)
{
int i,j;
int temp[n][n];
//判断旋转方向
for(i = 0; i < n; ++i)
for (j = 0; j < n;++j)
{
if(m == 1) //顺时针旋转90°
temp[j][n - i-1] = a[row + i][col + j];
else //逆时针旋转90°
temp[n-j-1][i] = a[row + i][col + j];
}
//替换a中对应旋转位置的数据
for(i=row;i<(row+n);i++)
for(j=col;j<(col+n);j++)
a[i][j]=temp[i-row][j-col];
}
编辑于 2018-02-11 22:03:01
回复(1)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()) {
int[][] arr = new int[5][5];
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
arr[i][j] = scanner.nextInt();
}
}
int a = scanner.nextInt();
int b = scanner.nextInt();
//对应下标为x-1,y-1
int x = scanner.nextInt();
int y = scanner.nextInt();
//获取以x，y为左上角的b²个数
int[][] arrBefore = new int[b][b];
for (int i = 0; i < b; i++) {
for (int j = 0; j < b; j++) {
arrBefore[i][j] = arr[i+x-1][j+y-1];
}
}
//将这些数按要求旋转
int[][] arrFlip = flip(arrBefore, a, b);
for (int i = 0; i < arrFlip.length; i++) {
for (int j = 0; j < arrFlip.length; j++) {
arr[i+x-1][j+y-1] = arrFlip[i][j];
}
}
//输出旋转后的数组
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
System.out.print(arr[i][j]+" ");
}
System.out.println();
}
}
}
public static int[][] flip(int[][] arrBefore,int a,int b) {
int[][] arrFlip = new int[b][b];
if (a == 1) { //顺时针翻转90度
for (int i = 0; i < b; i++) {
for (int j = 0; j < b; j++) {
arrFlip[i][j] = arrBefore[b-j-1][i];
}
}
return arrFlip;
}else if (a == 2) { //逆时针翻转90度
for (int i = 0; i < b; i++) {
for (int j = 0; j < b; j++) {
arrFlip[i][j] = arrBefore[j][b-i-1];
}
}
return arrFlip;
}
return arrFlip;
}
}
发表于 2024-03-27 20:56:03
回复(0)
#include <array>
#include <iostream>
#include <istream>
#include <ostream>
using namespace std;
using Matrix = array<array<int, 6>, 6>;
istream& operator>>(istream& in, Matrix& m) {
for (int i = 1; i <= 5; i++) {
for (int j = 1; j <= 5; j++) {
in >> m[i][j];
}
}
return in;
}
ostream& operator<<(ostream& out, const Matrix& m) {
for (int i = 1; i <= 5; i++) {
for (int j = 1; j <= 5; j++) {
out << m[i][j] << " ";
}
out << endl;
}
return out;
}
void turn(Matrix& m, int direction, int number, int x, int y) {
if (direction == 1 && number == 2) { //90度，顺时针，翻转4个数
int temp = m[x][y];
m[x][y] = m[x + 1][y];
m[x + 1][y] = m[x + 1][y + 1];
m[x + 1][y + 1] = m[x][y + 1];
m[x][y + 1] = temp;
} else if (direction == 1 && number == 3) { //90度，顺时针，翻转9个数
int temp = m[x][y];
m[x][y] = m[x + 2][y];
m[x + 2][y] = m[x + 2][y + 2];
m[x + 2][y + 2] = m[x][y + 2];
m[x][y + 2] = temp;
temp = m[x + 1][y];
m[x + 1][y] = m[x + 2][y + 1];
m[x + 2][y + 1] = m[x + 1][y + 2];
m[x + 1][y + 2] = m[x][y + 1];
m[x][y + 1] = temp;
} else if (direction == 2 && number == 2) { //90度，逆时针，翻转4个数
int temp = m[x][y];
m[x][y] = m[x][y + 1];
m[x][y + 1] = m[x + 1][y + 1];
m[x + 1][y + 1] = m[x + 1][y];
m[x + 1][y] = temp;
} else { //direction == 2 && number == 3（90度，逆时针，翻转9个数）
int temp = m[x][y];
m[x][y] = m[x][y + 2];
m[x][y + 2] = m[x + 2][y + 2];
m[x + 2][y + 2] = m[x + 2][y];
m[x + 2][y] = temp;
temp = m[x + 1][y];
m[x + 1][y] = m[x][y + 1];
m[x][y + 1] = m[x + 1][y + 2];
m[x + 1][y + 2] = m[x + 2][y + 1];
m[x + 2][y + 1] = temp;
}
}
int main() {
Matrix m;
int direction, number, x, y;
while (cin >> m >> direction >> number >> x >> y) {
turn(m, direction, number, x, y);
cout << m;
}
return 0;
}
发表于 2024-03-04 18:28:07
回复(0)
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
void turn(vector<vector<int>>& vv, int a, int b, int i, int j){
if (b == 2){
swap(vv[i][j], vv[i][j + 1]);
swap(vv[i + 1][j], vv[i + 1][j + 1]);
if (a == 1) swap(vv[i][j], vv[i + 1][j + 1]);
else swap(vv[i][j + 1], vv[i + 1][j]);
}else{
swap(vv[i][j], vv[i][j + 2]);
swap(vv[i + 2][j], vv[i + 2][j + 2]);
if (a == 1) swap(vv[i][j], vv[i + 2][j + 2]);
else swap(vv[i][j + 2], vv[i + 2][j]);
swap(vv[i][j + 1], vv[i + 1][j + 2]);
swap(vv[i + 1][j], vv[i + 2][j + 1]);
if (a == 1) swap(vv[i][j + 1], vv[i + 2][j + 1]);
else swap(vv[i + 1][j + 2], vv[i + 1][j]);
}
}
int main() {
vector<vector<int>> vv(5, vector<int>(5));
int a, b, i, j;
while (cin >> vv[0][0] >> vv[0][1] >> vv[0][2] >> vv[0][3] >> vv[0][4]){
for (int i = 1; i <= 4; ++i)
for (int j = 0; j <= 4; ++j)
cin >> vv[i][j];
cin >> a >> b >> i >> j;
turn(vv, a, b, i - 1, j - 1);
for (auto &v : vv){
for (int &n : v)
cout << n << " ";
cout << endl;
}
}
}
发表于 2023-03-09 22:53:07
回复(0)
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
//矩阵阶数
int n = 5;
struct Instruction{
vector<int> op;
pair<int, int> ad;
};
void input(vector<vector<int>>& Matrix) {
int num = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
cin >> num;
Matrix[i][j] = num;
}
}
}
//
void input(Instruction& I) {
int op1, op2;
cin >> op1 >> op2;
I.op.push_back(op1); I.op.push_back(op2);
cin >> I.ad.first >> I.ad.second;
}
//顺时针旋转二维的[(x, y), [px, py]]
void ROTATE(vector<vector<int>>& Matrix, int x, int y, int px, int py) {
int n = px - x;
vector<vector<int>> ans(Matrix);
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
ans[j + x][n - 1 - i + y] = Matrix[i + x][j + y];
}
}
Matrix = ans;
}
//逆时针等于转三次顺时针
void anti_clockWise(vector<vector<int>>& Matrix, int x, int y, int px, int py) {
for(int i = 0; i < 3; i++) {
ROTATE(Matrix, x, y, px, py);
}
}
//执行指令
void Excute(Instruction& I, vector<vector<int>>& Matrix) {
auto [x, y] = I.ad;
x--; y--;
auto px = x + I.op.at(1), py = y + I.op.at(1);
if(I.op.at(0) == 1) {
ROTATE(Matrix, x, y, px, py);
}else{
anti_clockWise(Matrix, x, y, px, py);
}
}
//打印
void printMatrix(const vector<vector<int>>& Matrix) {
for(auto row : Matrix) {
for(auto num : row) {
cout << num << " ";
}
cout << endl;
}
}
int main() {
//INIT
Instruction I;
vector<vector<int>> Matrix(n, vector<int>(n));
input(Matrix);
input(I);
Excute(I, Matrix);
printMatrix(Matrix);
}
发表于 2021-09-07 15:38:56
回复(0)
#include <cstdio>
int a[5][5]={0};
int temp[5][5]={0};
void change(int x,int y,int p,int b){
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
temp[i][j]=a[i][j];
}
}
if(x==1&&y==2){
for(int i=p+0;i<p+2;i++){
for(int j=b+0;j<b+2;j++){
temp[i][j]=a[j-b+p+2*b+1-2*j][i-p+b];
}
}
}
if(x==1&&y==3){
for(int i=p+0;i<p+3;i++){
for(int j=b+0;j<b+3;j++){
temp[i][j]=a[j-b+p+2*b+2-2*j][i-p+b];
}
}
}
if(x==2&&y==2){
for(int i=p+0;i<p+2;i++){
for(int j=b+0;j<b+2;j++){
temp[i][j]=a[j-b+p][i-p+b+2*p+1-2*i];
}
}
}
if(x==2&&y==3){
for(int i=p+0;i<p+3;i++){
for(int j=b+0;j<b+3;j++){
temp[i][j]=a[j-b+p][i-p+b+2*p+2-2*i];
}
}
}
}
int main(){
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
scanf("%d",&a[i][j]);
}
}
int x,y,p,b;
scanf("%d%d%d%d",&x,&y,&p,&b);
p--;
b--;
change(x,y,p,b);
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
if(j!=4) printf("%d ",temp[i][j]);
else printf("%d\n",temp[i][j]);
}
}
return 0;
}
int a[5][5]={0};
int temp[5][5]={0};
void change(int x,int y,int p,int b){
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
temp[i][j]=a[i][j];
}
}
if(x==1&&y==2){
for(int i=p+0;i<p+2;i++){
for(int j=b+0;j<b+2;j++){
temp[i][j]=a[j-b+p+2*b+1-2*j][i-p+b];
}
}
}
if(x==1&&y==3){
for(int i=p+0;i<p+3;i++){
for(int j=b+0;j<b+3;j++){
temp[i][j]=a[j-b+p+2*b+2-2*j][i-p+b];
}
}
}
if(x==2&&y==2){
for(int i=p+0;i<p+2;i++){
for(int j=b+0;j<b+2;j++){
temp[i][j]=a[j-b+p][i-p+b+2*p+1-2*i];
}
}
}
if(x==2&&y==3){
for(int i=p+0;i<p+3;i++){
for(int j=b+0;j<b+3;j++){
temp[i][j]=a[j-b+p][i-p+b+2*p+2-2*i];
}
}
}
}
int main(){
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
scanf("%d",&a[i][j]);
}
}
int x,y,p,b;
scanf("%d%d%d%d",&x,&y,&p,&b);
p--;
b--;
change(x,y,p,b);
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
if(j!=4) printf("%d ",temp[i][j]);
else printf("%d\n",temp[i][j]);
}
}
return 0;
}
发表于 2021-03-11 18:01:35
回复(0)
#include<iostream>
using namespace std;
int main()
{
int d[5][5],od[5][5],t1,t2,x,y;
while(cin >> d[0][0] >> d[0][1] >> d[0][2] >> d[0][3] >> d[0][4])
{
for(int i = 0;i < 5;i++) od[0][i] = d[0][i];
for(int i = 1;i < 5;i++)
{
for(int j = 0;j < 5;j++)
{
cin >> d[i][j];
od[i][j] = d[i][j];
}
}
cin >> t1 >> t2 >> x >> y;
if(t1 == 1 && t2 == 2)
{
for(int i = x - 1,m = y - 1;i < x + 1,m < y + 1;i++,m++)
{
for(int j = y - 1,n = x;j < y + 1,n >= x - 1;j++,n--)
{
od[i][j] = d[n][m];
}
}
}
else if(t1 == 1 && t2 == 3)
{
for(int i = x - 1,m = y - 1;i < x + 2,m < y + 2;i++,m++)
{
for(int j = y - 1,n = x + 1;j < y + 2,n >= x - 1;j++,n--)
{
od[i][j] = d[n][m];
}
}
}
else if(t1 == 2 && t2 == 2)
{
for(int i = x - 1,m = y;i < x + 1,m >= y - 1;i++,m--)
{
for(int j = y - 1,n = x - 1;j < y + 1,n < x + 1;j++,n++)
{
od[i][j] = d[n][m];
}
}
}
else if(t1 == 2 && t2 == 3)
{
for(int i = x - 1,m = y + 1;i < x + 2,m >= y - 1;i++,m--)
{
for(int j = y - 1,n = x - 1;j < y + 2,n < x + 2;j++,n++)
{
od[i][j] = d[n][m];
}
}
}
for(int i = 0;i < 5;i++)
{
for(int j = 0;j < 5;j++)
{
cout << od[i][j] << " ";
}
cout << endl;
}
}
return 0;
}
发表于 2021-02-22 20:10:38
回复(0)
#include<stdio.h>
#include<iostream>
#include<cstdio>
using namespace std;
int buf[5][5];
int helper[5][5];
void rotateRight(int matrix[5][5], int n) {
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = tmp;
}
}
for (int i = 0; i < n; i++) {
int left = 0, right = n - 1;
while (left < right) {
int tmp = matrix[i][left];
matrix[i][left] = matrix[i][right];
matrix[i][right] = tmp;
left++; right--;
}
}
}
void rotateLeft(int matrix[5][5], int n) {
rotateRight(matrix, n);
rotateRight(matrix, n);
rotateRight(matrix, n);
}
void printMatrix(int buf[5][5]) {
for (int i = 0; i < 5; i++) {
cout << buf[i][0] << " "<< buf[i][1] << " " << buf[i][2] << " " << buf[i][3] << " " << buf[i][4] << endl;
}
}
void solve(int matrix[5][5], int a, int b, int row, int col) {
for (int i = row; i < row + b; i++) {
for (int j = col; j < col + b; j++) {
helper[i - row][j - col] = matrix[i][j];
}
}
if (a == 1) rotateRight(helper, b);
else rotateLeft(helper, b);
for (int i = row; i < row + b; i++) {
for (int j = col; j < col + b; j++) {
matrix[i][j] = helper[i - row][j - col];
}
}
}
int main() {
int n;
int a, b, row, col;
while (scanf("%d", &n) != EOF) {
buf[0][0] = n;
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
if (i == 0 && j == 0) continue;
cin >> buf[i][j];
}
}
cin >> a >> b >> row >> col;
solve(buf, a, b, row - 1, col - 1);
printMatrix(buf);
}
return 0;
} 为了避免思路混乱，采用 helper 将需要进行旋转的子矩阵块提取出来，旋转好后再放回去。
发表于 2020-06-17 18:27:06
回复(0)
#include<cstdio>
#include<algorithm>
int v[6][6];
int main(){
int n1,n2,x,y;
while(scanf("%d",&v[1][1])!=EOF){
for(int i=1;i<=5;++i){
for(int j=1;j<=5;++j){
if(i!=1||j!=1)
scanf("%d",&v[i][j]);
}
}
scanf("%d %d %d %d",&n1,&n2,&x,&y);
for(int i=1;i<=5;++i){
for(int j=1;j<=5;++j){
if(i>=x&&i<x+n2&&j>=y&&j<y+n2){
int X1=i-x+1,Y1=j-y+1;
if(n1==1)
{
if(j==1)
printf("%d",v[x+n2-Y1][y+X1-1]);
else
printf(" %d",v[x+n2-Y1][y+X1-1]);
}
else{
if(j==1)
printf("%d",v[x+Y1-1][y+n2-X1]);
else
printf(" %d",v[x+Y1-1][y+n2-X1]);
}
}else
{
if(j==1)
printf("%d",v[i][j]);
else
printf(" %d",v[i][j]);
}
}
printf("\n");
}
}
return 0;
}
编辑于 2020-04-27 18:21:59
回复(0)
//用矩阵乘法实现的矩阵旋转，要思考的地方比较多，感觉暴力写可能更快吧
//要实现矩阵顺时针旋转90度，只要把矩阵转置再右乘一个单位矩阵的变形阵即可
//例如矩阵 1 2 要变成 3 1，可以先转置成 1 3再右乘 0 1即可
// 3 4 4 2 2 4 1 0
//要逆时针旋转90度只要左乘那个变形的单位阵即可
//左行右列原理
#include<iostream>
(720)#include<cstdio>
#include<algorithm>
(831)#include<cstring>
using namespace std;
const int MAXN = 6;
//写矩阵算法要特别注意i和j，千万不要错写！！！
//这题矩阵旋转我是按照矩阵乘法算的，思路正确
struct Matrix {
int row, col;
int matrix[MAXN][MAXN];
Matrix(int x,int y):row(x),col(y){ memset(matrix, 0, sizeof(matrix)); }
Matrix() { memset(matrix, 0, sizeof(matrix)); }
};
Matrix tansposition(Matrix a) {
Matrix answer = a;
for (int i = 0; i < a.row; i++) {
for (int j = 0; j < a.col; j++) {
answer.matrix[j][i] = a.matrix[i][j];
}
}
return answer;
}
Matrix ClockWise90Degree(int n) {
Matrix answer(n, n);
for (int i = 0; i < answer.row; i++) {
for (int j = 0; j < answer.col; j++) {
if (i + j == n - 1) {
answer.matrix[i][j] = 1;
}
else {
answer.matrix[i][j] = 0;
}
}
}
return answer;
}
//逆时针的矩阵似乎没必要，顺时针矩阵右乘欲旋转的矩阵即可
Matrix AntiClockWise90Degree(int n) {
Matrix answer(n, n);
for (int i = 0; i < answer.row; i++) {
for (int j = 0; j < answer.col; j++) {
if (i + j == n - 1) {
answer.matrix[i][j] = 1;
}
else {
answer.matrix[i][j] = 0;
}
}
}
return answer;
}
//默认a,b均为方阵
Matrix Multiple(Matrix a, Matrix b) {
Matrix answer(a.row, b.col);
memset(answer.matrix, 0, sizeof(answer.matrix));
for (int i = 0; i < a.row; i++) {
for (int j = 0; j < b.col; j++) {
for (int k = 0; k < a.col; k++) {
answer.matrix[i][j] += a.matrix[i][k] * b.matrix[k][j];
//cout<< a.matrix[i][k] <<endl;
//cout<< b.matrix[k][j] <<endl;
//cout<< answer.matrix[i][j] <<endl;
}
}
}
return answer;
}
//输出矩阵的算法
void print(Matrix a) {
for (int i = 0; i < a.row; i++) {
for (int j = 0; j < a.col; j++) {
if (j == 0) {
printf("%d", a.matrix[i][j]);
}
else {
printf(" %d", a.matrix[i][j]);
}
}
printf("\n");
}
return;
}
int main() {
Matrix res(5, 5);
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
scanf("%d", &res.matrix[i][j]);
}
}
int a, b, x, y;
Matrix temp;
scanf("%d%d%d%d", &a, &b, &x, &y);
//题目默认矩阵行，列均从1开始
x--;
y--;
if (b == 2) {
temp = ClockWise90Degree(2);
Matrix t(2, 2);
for (int i = x; i < x + 2; i++) {
for (int j = y; j < y + 2; j++) {
t.matrix[i - x][j - y] = res.matrix[i][j];
}
}
t = tansposition(t);
//print(temp);
//print(t);
if (a == 1) {
temp = Multiple(t, temp);
for (int i = x; i < x + 2; i++) {
for (int j = y; j < y + 2; j++) {
res.matrix[i][j] = temp.matrix[i - x][j - y];
}
}
}
else {
temp = Multiple(temp, t);
for (int i = x; i < x + 2; i++) {
for (int j = y; j < y + 2; j++) {
res.matrix[i][j] = temp.matrix[i - x][j - y];
}
}
}
}
else {
temp = ClockWise90Degree(3);
Matrix t(3, 3);
for (int i = x; i < x + 3; i++) {
for (int j = y; j < y + 3; j++) {
t.matrix[i - x][j - y] = res.matrix[i][j];
}
}
t = tansposition(t);
if (a == 1) {
temp = Multiple(t, temp);
for (int i = x; i < x + 3; i++) {
for (int j = y; j < y + 3; j++) {
res.matrix[i][j] = temp.matrix[i - x][j - y];
}
}
}
else {
temp = Multiple(temp, t);
for (int i = x; i < x + 3; i++) {
for (int j = y; j < y + 3; j++) {
res.matrix[i][j] = temp.matrix[i - x][j - y];
}
}
}
}
print(res);
return 0;
}
编辑于 2020-04-20 17:43:30
回复(0)
while True:
try:
c = []
for _ in range(5):
c.append(list(input().split()))
b = list(map(int,input().split()))
x = b[2] - 1
y = b[3] - 1
if b[0] == 1 and b[1] == 2:
c[x][y],c[x][y+1],c[x+1][y+1],c[x+1][y] = c[x+1][y],c[x][y],c[x][y+1],c[x+1][y+1]
elif b[0] == 1 and b[1] == 3:
c[x][y],c[x][y+1],c[x][y+2],c[x+1][y+2],c[x+2][y+2],c[x+2][y+1],c[x+2][y],c[x+1][y]\
= c[x+2][y],c[x+1][y],c[x][y],c[x][y+1],c[x][y+2],c[x+1][y+2],c[x+2][y+2],c[x+2][y+1]
elif b[0] == 2 and b[1] == 2:
c[x][y],c[x][y+1],c[x+1][y+1],c[x+1][y] = c[x][y+1],c[x+1][y+1],c[x+1][y],c[x][y]
else:
c[x][y],c[x][y+1],c[x][y+2],c[x+1][y+2],c[x+2][y+2],c[x+2][y+1],c[x+2][y],c[x+1][y]\
= c[x][y+2],c[x+1][y+2],c[x+2][y+2],c[x+2][y+1],c[x+2][y],c[x+1][y],c[x][y],c[x][y+1]
for i in c:
print(' '.join(map(str,i)))
except:
break
发表于 2020-04-13 12:03:37
回复(0)
#include<iostream>
(720)#include<vector>
using namespace std;
int main()
{
vector<vector<int>> martix(5);
for (int i = 0; i < 5; i++)
martix[i].resize(5);
for (int i = 0; i < 5; i++)
for (int j = 0; j < 5; j++)
cin >> martix[i][j];
vector<vector<int>> temp = martix;
int a, b, x, y;
cin >> a >> b >> x >> y;
y--;
x--;
if (a == 1)
for (int i = x; i < x + b; i++)
for (int j = y; j < y + b; j++)
martix[j - y + x][y + b - 1 - i + x] = temp[i][j];
else
for (int i = x; i < x + b; i++)
for (int j = y; j < y + b; j++)
martix[b - j + y + x - 1][i - x + y] = temp[i][j];
for (int i = 0; i < 5; i++)
{
for (int j = 0; j < 5; j++)
{
cout << martix[i][j] << " ";
}
cout << endl;
}
}
发表于 2020-04-11 12:23:25
回复(0)
逆时针旋转就是顺时针旋转三次。
#include<iostream>
(720)#include<cstring>
using namespace std;
int tmp[3][3];
void rotate(int a[][5],int n,int x,int y){
for(int i =0;i<n;i++)
for(int j=0;j<n;j++)
tmp[j][n-i-1] = a[i+x-1][j+y-1];
for(int i =0;i<n;i++)
for(int j=0;j<n;j++)
a[i+x-1][j+y-1] = tmp[i][j];
}
int main(){
int matrix[5][5];
int a[4];
for(int i =0;i<5;i++)
for(int j=0;j<5;j++)
cin>>matrix[i][j];
for(int i = 0;i<4;i++)
cin>>a[i];
if(a[0]==1&&a[1]==2){
rotate(matrix,2,a[2],a[3]);
}
if(a[0]==1&&a[1]==3){
rotate(matrix,3,a[2],a[3]);
}
if(a[0]==2&&a[1]==2){
rotate(matrix,2,a[2],a[3]);
rotate(matrix,2,a[2],a[3]);
rotate(matrix,2,a[2],a[3]);
}
if(a[0]==2&&a[1]==3){
rotate(matrix,3,a[2],a[3]);
rotate(matrix,3,a[2],a[3]);
rotate(matrix,3,a[2],a[3]);
}
for(int i =0;i<5;i++){
for(int j=0;j<5;j++)
cout<<matrix[i][j]<<" ";
cout<<endl;
}
}
发表于 2020-02-25 16:29:02
回复(0)
#include<iostream>
using namespace std;
int a[5][5];
int res[5][5];
typedef struct vec{
double x,y;
vec(){
x=0;y=0;
}
vec(double a,double b):x(a),y(b){
}
vec operator-(const vec& v){
return vec(x-v.x,y-v.y);
}
vec operator+(const vec& v){
return vec(x+v.x,y+v.y);
}
vec rot(int op){
if(op==1)
return vec(-y,x);
else return vec(y,-x);
}
}vec;
int main(){
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
cin>>a[i][j];
res[i][j]=a[i][j];
}
}
int o1,o2,x,y;
cin>>o1>>o2>>x>>y;
for(int i=x-1;i<x-1+o2;i++){
for(int j=y-1;j<y-1+o2;j++){
vec O;
O.x=((double)o2-1)/2+y-1;
O.y=((double)o2-1)/2+x-1;
vec des=(vec(j,i)-O).rot(o1)+O;
res[(int)des.y][(int)des.x]=a[i][j];
}
}
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
cout<<res[i][j]<<' ';
}
cout<<endl;
}
return 0;
}
利用线性代数的知识，可以直接用旋转矩阵乘以坐标得到结果
发表于 2019-09-18 22:26:27
回复(0)
def one(x,y,index):
list1=[]
for j in range(y-1,y-1+index):
re=[]
for i in range(x-1,x-1+index):
re.append(arr[i][j])
list1.append(re[::-1])
for i in range(x-1,x-1+index):
for j in range(y-1,y-1+index):
arr[i][j]=list1[i-x+1][j-y+1]
def two(x,y,index):
list1=[]
for j in range(y-2+index,y-2,-1):
re=[]
for i in range(x-1,x-1+index):
re.append(arr[i][j])
list1.append(re)
for i in range(x-1,x-1+index):
for j in range(y-1,y-1+index):
arr[i][j]=list1[i-x+1][j-y+1]
while True:
try:
arr=[]
for i in range(5):
arr.append(list(input().strip().split(' ')))
inp=list(map(int,input().strip().split(' ')))
if [inp[0],inp[1]]==[1,2]:
one(inp[2],inp[3],2)
elif [inp[0],inp[1]]==[1,3]:
one(inp[2],inp[3],3)
elif [inp[0],inp[1]]==[2,2]:
two(inp[2],inp[3],2)
else:
two(inp[2],inp[3],3)
for i in arr:
print(' '.join(i))
except:
break
发表于 2019-08-31 22:42:15
回复(0)
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