[编程题]旋转矩阵
- 热度指数：5626 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
任意输入两个9阶以下矩阵，要求判断第二个是否是第一个的旋转矩阵，如果是，输出旋转角度（0、90、180、270），如果不是，输出-1。
要求先输入矩阵阶数，然后输入两个矩阵，每行两个数之间可以用任意个空格分隔。行之间用回车分隔，两个矩阵间用任意的回车分隔。
输入描述:
输入有多组数据。 每组数据第一行输入n(1<=n<=9)，从第二行开始输入两个n阶矩阵。
输出描述:
判断第二个是否是第一个的旋转矩阵，如果是，输出旋转角度（0、90、180、270），如果不是，输出-1。 如果旋转角度的结果有多个，则输出最小的那个。
示例1
输入
3 1 2 3 4 5 6 7 8 9 7 4 1 8 5 2 9 6 3
输出
90
测试数据比较亲切
#include <cstdio>
int a[9][9], b[9][9];
int main () {
int n;
scanf("%d", &n);
for (int i=0; i<n; ++i)
for (int j=0; j<n; ++j)
scanf("%d", &a[i][j]);
for (int i=0; i<n; ++i)
for (int j=0; j<n; ++j)
scanf("%d", &b[i][j]);
if (**a == **b)
printf("0");
else if (a[1][2] == b[2][n-2])
printf("90");
else if (a[1][2] == b[n-2][n-3])
printf("180");
else if (a[1][2] == b[n-3][1])
printf("270");
else
printf("-1");
} （钻空子的代码）
编辑于 2020-02-15 00:31:43
回复(3)
//一楼的答案真的很简洁了
#include<iostream>
using namespace std;
int main() {
int n;
while (cin >> n) {
int **a = new int*[n], **b = new int*[n], **c = new int*[n],**d = new int*[n],**e = new int*[n];
for (int i = 0; i<n; i++) {
a[i] = new int[n];
b[i] = new int[n];
c[i] = new int[n];
d[i] = new int[n];
e[i] = new int[n];
}
for (int i = 0; i<n; i++)
for (int j = 0; j<n; j++)
cin >> a[i][j];
for (int i = 0; i<n; i++)
for (int j = 0; j<n; j++)
cin >> e[i][j];
for (int i = 0; i<n; i++)
for (int j = n - 1; j >= 0; j--)
b[i][n - 1 - j] = a[j][i];
for (int i = 0; i<n; i++)
for (int j = n - 1; j >= 0; j--)
c[i][n - 1 - j] = b[j][i];
for (int i = 0; i<n; i++)
for (int j = n - 1; j >= 0; j--)
d[i][n - 1 - j] = c[j][i];
bool judge = 1;
for (int i = 0; i < n&&judge; i++)
for (int j = 0; j < n; j++)
if (a[i][j] != e[i][j]) {
judge = 0;
break;
}
if (judge)
cout << "0" << endl;
else {
judge = 1;
for (int i = 0; i < n&&judge; i++)
for (int j = 0; j < n; j++)
if (b[i][j] != e[i][j]) {
judge = 0;
break;
}
if (judge)
cout << "90" << endl;
else {
judge = 1;
for (int i = 0; i < n&&judge; i++)
for (int j = 0; j < n; j++)
if (c[i][j] != e[i][j]) {
judge = 0;
break;
}
if (judge)
cout << "180" << endl;
else {
judge = 1;
for (int i = 0; i < n&&judge; i++)
for (int j = 0; j < n; j++)
if (d[i][j] != e[i][j]) {
judge = 0;
break;
}
if (judge)
cout << "270" << endl;
else
cout << "-1" << endl;
}
}
}
}
}
发表于 2020-01-14 10:19:37
回复(0)
通过这道题目，我学会了new二维数组以及-1要用双引号
。。。
好了我知道丢人了
#include<iostream>
using namespace std;
int n;
int **rot(int a[9][9], int angle){
int **b = new int*[9];
for(int i = 0; i < 9; i++){
b[i] = new int[9];
}
if(angle == 1){
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
b[j][n - i - 1] = a[i][j];
}
}
}
else if(angle == 2){
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
b[n - i - 1][n - j - 1] = a[i][j];
}
}
}
else if(angle == 3){
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
b[n - j - 1][i] = a[i][j];
}
}
}
else{
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
b[i][j] = a[i][j];
}
}
}
return b;
}
bool mysame(int a[9][9], int **b){
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(a[i][j] != b[i][j]){
return false;
}
}
}
return true;
}
int main(){
int a[9][9], b[9][9];
while(cin >> n){
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
cin >> a[i][j];
}
}
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
cin >> b[i][j];
}
}
bool flag = false, temp;
for(int angle = 0; angle < 4; angle++){
temp = mysame(b, rot(a, angle));
flag = flag || temp;
if(temp == true){
cout << 90 * angle << endl;
break;
}
}
if(flag == false){
cout << "-1" << endl;
}
}
return 0;
}
发表于 2019-04-10 16:08:35
回复(0)
#include<iostream>
using namespace std;
int test1163()
{
int n;
while(cin>>n)
{
bool v0=true,v90=true,v180=true,v270=true;
int *m1=new int[n*n];
int *m2=new int[n*n];
int *m90=new int[n*n];
int *m180=new int[n*n];
int *m270=new int[n*n];
for(int i=0;i<n*n;i++)
{
cin>>m1[i];
}
for(int j=0;j<n*n;j++)
{
cin>>m2[j];
}
for(int a=0;a<n;a++)
{
for(int j=0;j<n;j++)
{
m90[n*a+j]=m1[n*(n-1-j)+a];
m180[n*a+j]=m1[n*(n-1-a)+(n-1-j)];
m270[n*a+j]=m1[n*j+(n-1-a)];
}
}
for(int b=0;b<n*n;b++)
{
if(m2[b]!=m1[b])
v0=false;
if(m2[b]!=m90[b])
v90=false;
if(m2[b]!=m180[b])
v180=false;
if(m2[b]!=m270[b])
v270=false;
}
if(v0)
cout<<"0"<<endl;
else if(v90)
cout<<"90"<<endl;
else if(v180)
cout<<"180"<<endl;
else if(v270)
cout<<"270"<<endl;
else
cout<<"-1"<<endl;
delete []m1;
delete []m2;
delete []m90;
delete []m180;
delete []m270;
}
return 0;
}
int main()
{
test1163();
return 0;
}
发表于 2017-02-28 12:12:41
回复(0)
解法一：
#include <iostream>
using namespace std;
int main() {
int N;
while (cin >> N) {
int A[N][N], B[N][N], flag[4] = { 0 };
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
cin >> A[i][j];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
cin >> B[i][j];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
if (A[i][j] != B[i][j])
flag[0] = 1;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
if (A[i][j] != B[j][N - i - 1])
flag[1] = 1;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
if (A[i][j] != B[N - i - 1][N - j - 1])
flag[2] = 1;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
if (A[i][j] != B[N - j - 1][i])
flag[3] = 1;
if (flag[0] == 0)
cout << "0" << endl;
else if (flag[1] == 0)
cout << "90" << endl;
else if (flag[2] == 0)
cout << "180" << endl;
else if (flag[3] == 0)
cout << "270" << endl;
else
cout << -1 << endl;
}
return 0;
}
解法二：
#include <iostream>
using namespace std;
int main() {
int N;
while (cin >> N) {
int A[N][N], B[N][N], result = 0;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
cin >> A[i][j];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
cin >> B[i][j];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++) {
if (result == 0 && A[i][j] == B[i][j])
continue;
else {
result = 90;
if (result == 90 && (A[i][j] == B[j][N - i - 1]))
continue;
else {
result = 180;
if (result == 180 && (A[i][j] == B[N - i - 1][N - j - 1]))
continue;
else {
result = 270;
if (result == 270 && (A[i][j] == B[N - j - 1][i]))
continue;
else {
result = -1;
break;
}
}
}
}
}
cout << result << endl;
}
}
编辑于 2017-02-14 17:42:28
回复(0)
#include<stdio.h>
void turn90(int A[10][10],int n){ //将A顺时针旋转90度后存储在A中
int B[10][10];
for(int i=0;i<n;i++){ //讲矩阵A顺时针旋转90度后暂存在B矩阵中
for(int j=0;j<n;j++){
B[i][j]=A[n-1-j][i];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
A[i][j]=B[i][j];
}
}
}
int judge(int A[10][10],int B[10][10],int n){ //判断A，B两个矩阵是否相同
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(A[i][j]!=B[i][j])
return 0;
}
}
return 1; //如果相同则返回1；
}
int main(){
int n;
while(scanf("%d",&n)!=EOF){
int buf_A[10][10];
int buf_B[10][10];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
scanf("%d",&buf_A[i][j]);
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
scanf("%d",&buf_B[i][j]);
}
}
int tmp[10][10];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
tmp[i][j]=buf_A[i][j];
}
}
int cnt=0;
for(int i=1;i<=4;i++){
int m=judge(buf_A,buf_B,n);
if(m==1) {
cnt=i;
break;
}
turn90(buf_A,n); //将buf_A顺时针旋转90度放在tmp中
}
if(cnt==0) printf("-1");
else if(cnt==1) printf("0");
else if(cnt==2) printf("90");
else if(cnt==3) printf("180");
else if(cnt==4) printf("270");
}
return 0;
}
发表于 2018-03-22 22:37:04
回复(0)
#include<stdio.h>
int main()
{
int n;
int i,j;
while(scanf("%d",&n)!=EOF)
{
int a[9][9],b[9][9];
for(i=0;i<n;i++) /**录入矩阵的值*/
for(j=0;j<n;j++)
scanf("%d",&a[i][j]);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&b[i][j]);
int flag[4]={1,1,1,1}; /**标记0,90,180,270四个方向*/
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
if(a[i][j]!=b[i][j]) flag[0]=0;
if(a[n-1-j][i]!=b[i][j]) flag[1]=0;
if(a[n-1-i][n-1-j]!=b[i][j]) flag[2]=0;
if(a[j][n-1-i]!=b[i][j]) flag[3]=0;
}
i=0;
while(flag[i]==0) i++;
if(i==4)
printf("-1\n");
else
printf("%d\n",90*i);
}
return 0;
}
发表于 2017-03-01 20:49:57
回复(7)
//思路和之前一样，矩阵的旋转等同于先上下翻转，再沿着对角线对称。
#include <cstdio>
(802)#include <iostream>
#include <algorithm>
using namespace std;
// 顺时针旋转90度
void rotate(int (*a)[10], int len)
{
// 先上下翻转
int top = 0, down = len - 1;
while (top < down) {
for (int i = 0; i < len; ++i) {
swap(a[top][i], a[down][i]);
}
top++, down--;
}
// 沿着对角线翻转
for (int i = 0; i < len; ++i) {
for (int j = i+1; j < len; ++j) {
swap(a[i][j], a[j][i]);
}
}
}
// 判断是否相同
bool isSame(int a1[10][10], int a2[10][10], int n) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if(a1[i][j] != a2[i][j])
return false;
}
}
return true;
}
int arr1[10][10];
int arr2[10][10];
int main()
{
freopen("data.txt", "r", stdin);
int n;
scanf("%d", &n);
int i, j;
for (i = 0; i < n; ++i) {
for (j = 0; j < n; ++j) {
scanf("%d", &arr1[i][j]);
}
}
for (i = 0; i < n; ++i) {
for (j = 0; j < n; ++j) {
scanf("%d", &arr2[i][j]);
}
}
int sec = 0;
if(isSame(arr1, arr2, n)) {
printf("%d", sec);
return 0;
}
sec+=90;
rotate(arr1, n);
if(isSame(arr1, arr2, n)) {
printf("%d", sec);
return 0;
}
sec+=90;
rotate(arr1, n);
if(isSame(arr1, arr2, n)) {
printf("%d", sec);
return 0;
}
sec+=90;
rotate(arr1, n);
if(isSame(arr1, arr2, n)) {
printf("%d", sec);
return 0;
}
printf("-1");
return 0;
}
发表于 2020-03-11 19:34:51
回复(0)
看起来长，实际很多一样的，改一点就行
#include <bits/stdc++.h>
using namespace std;
bool is_90(int a[9][9],int b[9][9])
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
if(a[i][j]!=b[j][n-1-i])
{
return false;
}
}
return true;
}
bool is_180(int a[9][9],int b[9][9])
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
if(a[i][j]!=b[n-1-i][n-1-j])
{
return false;
}
}
return true;
}
bool is_equal(int a[9][9],int b[9][9])
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
if(a[i][j]!=b[i][j])
{
return false;
}
}
return true;
}
int main()
{
int n;
while(cin>>n)
{
int a[9][9],b[9][9];
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
cin>>a[i][j];
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
cin>>b[i][j];
if(is_equal(a,b))
{
cout<<0<<endl;
continue;
}
if(is_90(a,b))
{
cout<<90<<endl;
continue;
}
if(is_180(a,b))
{
cout<<180<<endl;
continue;
}
if(is_90(b,a))//a转270度得b，等价于b转90度得a
{
cout<<270<<endl;
continue;
}
cout<<-1<<endl;
}
return 0;
}
编辑于 2024-03-20 21:31:43
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()) {
int n = scanner.nextInt();
int[][] arr = new int[n][n];
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++) {
arr[i][j] = scanner.nextInt();
}
}
//旋转90度
int[][] arr90 = new int[n][n];
for (int i = 0; i < arr90.length; i++) {
for (int j = 0; j < arr90[i].length; j++) {
arr90[i][j] = arr[n-j-1][i];
}
}
int[][] arr180 = new int[n][n];
for (int i = 0; i < arr180.length; i++) {
for (int j = 0; j < arr180[i].length; j++) {
arr180[i][j] = arr90[n-j-1][i];
}
}
int[][] arr270 = new int[n][n];
for (int i = 0; i < arr270.length; i++) {
for (int j = 0; j < arr270[i].length; j++) {
arr270[i][j] = arr180[n-j-1][i];
}
}
int[][] arrLast = new int[n][n];
for (int i = 0; i < arrLast.length; i++) {
for (int j = 0; j < arrLast[i].length; j++) {
arrLast[i][j] = scanner.nextInt();
}
}
if (isEqual(arrLast, arr)) {
System.out.println(0);
}else if (isEqual(arrLast, arr90)) {
System.out.println(90);
}else if (isEqual(arrLast, arr180)) {
System.out.println(180);
}else if (isEqual(arrLast, arr270)) {
System.out.println(270);
}else {
System.out.println(-1);
}
}
}
public static boolean isEqual(int[][] arr,int[][] arr2) {
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++) {
if (arr[i][j] != arr2[i][j]) {
return false;
}
}
}
return true;
}
}
编辑于 2024-03-25 20:06:23
回复(0)
第一次写的时候转反了。。。
还是封装了一下
#include<stdio.h>
int arr1[100][100];
int arr2[100][100];
bool IsSame(int N){
int flag = 1;
for(int i=1;i<=N;i++){
for(int j=1;j<=N;j++){
if(arr1[i][j]!=arr2[i][j]){
return false;
}
}
}
return true;
}
void Switch(int N){
int tmp[100][100];
for(int i=1;i<=N;i++){
for(int j=N;j>0;j--){
tmp[i][N-j+1] = arr1[j][i];
}
}
for(int i=1;i<=N;i++){
for(int j=1;j<=N;j++){
arr1[i][j] = tmp[i][j];
}
}
}
int main(){
int N;
scanf("%d",&N);
for(int i=1;i<=N;i++){
for(int j=1;j<=N;j++){
scanf("%d",&arr1[i][j]);
}
}
for(int i=1;i<=N;i++){
for(int j=1;j<=N;j++){
scanf("%d",&arr2[i][j]);
}
}
int count = 0;
while(!IsSame(N)){
Switch(N);
count+=90;
if(count==360) break;
}
if(count<360) printf("%d\n",count);
else printf("-1\n");
}
编辑于 2024-03-18 20:42:15
回复(0)
原地旋转
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
void rotate(vector<vector<int>>& vv){
int n = vv.size();
int row = n / 2;
int col = (n % 2 == 0) ? (n / 2) : (n / 2 + 1);
for (int i = 0; i < row; ++i)
for (int j = 0; j < col; ++j){
swap(vv[i][j], vv[j][n-i-1]);
swap(vv[n-i-1][n-j-1], vv[n-j-1][i]);
swap(vv[i][j], vv[n-i-1][n-j-1]);
}
}
bool compare(const vector<vector<int>>& vv1, const vector<vector<int>>& vv2){
int n = vv1.size();
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
if (vv1[i][j] != vv2[i][j]) return false;
return true;
}
int getRotateAngle(vector<vector<int>>& vv1, const vector<vector<int>>& vv2){
for (int i = 1; i <= 3; ++i){
rotate(vv1);
if (compare(vv1, vv2)) return i * 90;
}
return -1;
}
int main() {
int n;
while (cin >> n){
vector<vector<int>> vv1(n, vector<int>(n));
vector<vector<int>> vv2(n, vector<int>(n));
for (auto &v : vv1)
for (int &i : v)
cin >> i;
for (auto &v : vv2)
for (int &i : v)
cin >> i;
if (compare(vv1, vv2)){
cout << 0 << endl;
continue;
}
cout << getRotateAngle(vv1, vv2) << endl;
}
}
发表于 2023-03-05 22:54:11
回复(0)
#include <stdio.h>
#include <string.h>
using namespace std;
bool cmp(int *a, int *b, int len) {
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
// 利用数组的顺序存储从而访问数组元素
if (*(a + i * len + j) != *(b + i * len + j)) {
return false;
}
}
}
return true;
}
int main() {
int n;
scanf("%d", &n);
int mat_1[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &mat_1[i][j]);
}
}
int mat_2[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &mat_2[i][j]);
}
}
int degree = 0;
bool flag = false;
int temp[n][n];
// 依次旋转 0、90、180、270 度
for (int i = 0; i < 4; i++) {
if (!cmp(&mat_1[0][0], &mat_2[0][0], n)) {
// rotate
for (int r = 0; r < n; r++) {
for (int c = 0; c < n; c++) {
temp[c][n - 1 - r] = mat_1[r][c];
}
}
// 将旋转后的矩阵 temp 赋值给 mat_1
memcpy(mat_1, temp, sizeof(temp));
// 累加旋转的度数
degree += 90;
} else {
flag = true;
}
}
if (flag) printf("%d", degree);
else printf("-1");
}
发表于 2022-03-29 16:27:07
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#include<iostream>
#include<cstdio>
using namespace std;
const int N = 10;
int matrix1[N][N], matrix2[N][N];
int Judge(int n){
bool flag = true;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(matrix1[i][j] != matrix2[i][j]){
flag = false;
break;
}
}
if(!flag) break;
}
if(flag) return 0;
flag = true;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(matrix1[i][j] != matrix2[j][n - i - 1]){
flag = false;
break;
}
}
if(!flag) break;
}
if(flag) return 90;
flag = true;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(matrix1[i][j] != matrix2[n - i - 1][n - j - 1]){
flag = false;
break;
}
}
if(!flag) break;
}
if(flag) return 180;
flag = true;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(matrix1[i][j] != matrix2[n - j - 1][i]){
flag = false;
break;
}
}
if(!flag) break;
}
if(flag) return 270;
return -1;
}
int main(){
int n;
while(scanf("%d", &n) != EOF){
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
scanf("%d", &matrix1[i][j]);
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
scanf("%d", &matrix2[i][j]);
printf("%d\n", Judge(n));
}
return 0;
}
发表于 2022-03-04 21:15:00
回复(0)
#include <iostream>
#include <vector>
using namespace std;
int n = 0;
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 ROTATE(vector<vector<int>>& Matrix) {
vector<vector<int>> ans(n, vector<int>(n));
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
ans[j][n - 1 - i] = Matrix[i][j];
}
}
Matrix = ans;
}
bool isEqual(const vector<vector<int>>& A, const vector<vector<int>>& B) {
int m = A.size();
if(m == 0 || m != B.size()) {
return false;
}
int n = A[0].size();
if(n == 0 || n != B[0].size()) {
return false;
}
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(A[i][j] != B[i][j]) {
return false;
}
}
}
return true;
}
int main() {
//INIT
cin >> n;
vector<int> Angle{0,90,180,270,-1};
vector<vector<int>> MatrixA(n, vector<int>(n));
vector<vector<int>> MatrixB(n, vector<int>(n));
inPut(MatrixA); inPut(MatrixB);
for(int i = 0; i <= 4; i++) {
if(i == 4) {
cout << Angle[i];
}
if(isEqual(MatrixA, MatrixB)) {
cout << Angle[i];
break;
}
ROTATE(MatrixA);
}
return 0;
}
发表于 2021-09-07 14:48:21
回复(0)
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
int main(){
int n ;
while(cin>>n){
int ma[n][n], mb[n][n];
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++) cin>>ma[i][j];
}
//对ma进行转置运算
for(int i = 0; i < n; i++){
for(int j = i+1; j < n; j++) swap(ma[i][j], ma[j][i]);
}
int flag90 = 0, flag180 = 0, flag270 = 0, flag0=0;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
cin>>mb[i][j];
if(mb[i][j]==ma[j][i]) flag0++;
//90度与转置后的矩阵左右对称
if(mb[i][j]==ma[i][n-j-1]) flag90++;
if(mb[i][j]==ma[n-j-1][n-i-1]) flag180++;
//270度与转置后的矩阵上下对称
if(mb[i][j]==ma[n-i-1][j]) flag270++;
}
}
if(flag0==n*n||flag90==n*n||flag270==n*n||flag180==n*n){
int temp = flag0==n*n?0:(flag90==n*n?90:(flag180==n*n?180:270));
cout<<temp<<endl;
} else cout<<-1<<endl;
}
}
发表于 2021-03-20 10:12:24
回复(0)
#include <iostream>
using namespace std;
int arr1[10][10] = {0};
int arr2[10][10] = {0};
int temp[10][10] = {0};
void rev(int n){
int index = n - 1;
for(int i = 0;i < n;i++){
for(int j = 0;j < n;j++){
temp[j][index] = arr1[i][j];
}
index--;
}
for(int i = 0;i < n;i++){
for(int j = 0;j < n;j++){
arr1[i][j] = temp[i][j];
}
}
}
bool tojudge(int n){
for(int i = 0;i < n;i++){
for(int j = 0;j < n;j++){
if(arr1[i][j] != arr2[i][j])
return false;
}
}
return true;
}
int main(){
int n;
while(cin >> n){
for(int i = 0;i < n;i++){
for(int j = 0;j < n;j++){
cin >> arr1[i][j];
}
}
for(int i = 0;i < n;i++){
for(int j = 0;j < n;j++){
cin >> arr2[i][j];
}
}
if(tojudge(n)){
cout << "0" << endl;
continue;
}
bool f = false;
for(int i = 1;i <= 3;i++){
rev(n);
if(tojudge(n)){
f = true;
cout << i * 90 << endl;
break;
}
}
if(f == false)
cout << "-1" << endl;
}
}
发表于 2021-03-06 17:06:58
回复(0)
#include<iostream>
using namespace std;
int main(){
int n;
while(cin>>n){
int a[9][9],b[9][9];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>a[i][j];
}
}
bool o=true;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>b[i][j];
if(b[i][j]!=a[i][j])
o=false;
}
}
//90 第0行等于最后一列，第1行等于倒数第二列，最后一行等于第一列
bool ninty=true;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(a[i][j]!=b[j][n-i-1]){
ninty=false;
break;
}
}
}
//180 第0行等于最后一行倒着，第1行等于倒数第二行倒着，最后一行等于第一行倒着
bool oneeighto=true;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(a[i][j]!=b[n-i-1][n-j-1]){
oneeighto=false;
break;
}
}
}
//180 第0行等于第0列倒着，第1行等于第1列倒着
bool twoseveno=true;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(a[i][j]!=b[n-j-1][i]){
twoseveno=false;
break;
}
}
}
if(o==true){
cout<<0<<endl;
}else if(ninty==true){
cout<<90<<endl;
}else if(oneeighto==true){
cout<<180<<endl;
}else if(twoseveno==true){
cout<<270<<endl;
}else{
cout<<-1<<endl;
}
}
}
发表于 2020-07-14 00:46:29
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNextInt()){
int n = sc.nextInt();
int[][] arr1 = new int[n][n];
int[][] arr2 = new int[n][n];
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
arr1[i][j] = sc.nextInt();
}
}
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
arr2[i][j] = sc.nextInt();
}
}
whichAngle(arr1, arr2, n);
}
}
static int[][] rotate90(int[][] arr, int n){
int[][] rArr = new int[n][n];
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
rArr[i][j] = arr[n-1-j][i];
}
}
return rArr;
}
static boolean isEqual(int[][] arr1, int [][] arr2, int n){
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
if (arr1[i][j] != arr2[i][j]){
return false;
}
}
}
return true;
}
static void whichAngle(int[][] arr1, int[][] arr2, int n){
int[][] arr90 = rotate90(arr1, n);
int[][] arr180 = rotate90(arr90, n);
int[][] arr270 = rotate90(arr180, n);
if (isEqual(arr1, arr2, n)){
System.out.println("0");
return;
} else if (isEqual(arr90, arr2, n)){
System.out.println("90");
return;
} else if (isEqual(arr180, arr2, n)){
System.out.println("180");
return;
} else if (isEqual(arr270, arr2, n)){
System.out.println("270");
return;
} else {
System.out.println("-1");
return;
}
}
}
发表于 2020-05-11 11:01:23
回复(0)
#include<cstdio>
#include<vector>
using namespace std;
int main(){
int n;
while(scanf("%d",&n)!=EOF){
vector<vector<int>> v(n,vector<int>(n)),v1(n,vector<int>(n)),v2(n,vector<int>(n));
for(int i=0;i<n;++i){
for(int j=0;j<n;++j){
scanf("%d",&v[i][j]);
}
}
int flag=0;
for(int i=0;i<n;++i){
for(int j=0;j<n;++j){
scanf("%d",&v1[i][j]);
if(v[i][j]!=v1[i][j])
{
flag=1;
}
}
}
if(flag==0)
{
printf("0\n");
continue;
}
int k;
for(k=0;k<3;++k){
flag=0;
for(int i=0;i<n;++i){
for(int j=0;j<n;++j){
if(v[j][i]!=v1[i][n-j-1])
{
flag=1;
}
v2[i][n-j-1]=v[j][i];
}
}
if(flag==0){
printf("%d\n",(k+1)*90);
break;
}
v=v2;
}
if(k==3)
printf("-1\n");
}
return 0;
}
编辑于 2020-04-24 14:50:24
回复(0)
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