###### tags: `code`

# c14086028 homework-3

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define N 5
#define A 50
void ref(int ,int [],int []);
void rot(int ,int [],int []);
void refandrot(int,int[],int [],int[]);
void rxry(int,int,int[],int[],int[]);
int main() {
	int a[N] = { };
	int b[N] = { };
	int c[N] = { };
	int d[A][N] = { {} };
	int e[A][N] = { {} };
	/*give value of a[i],b[i],c[i]-b*/
	for (int i = 0; i < N; i++) {
		a[i] = i;
		b[i] = i;
		c[i] = i;
	}
	/*give value of a[i],b[i],c[i]-e*/

	/*print rot-k-b*/
	for (int k = 1; k <= N; k++) {
		rot(k,a,b);
		printf("rotation-%d\n", k);
		for (int i = 0; i < N; i++) {
			printf("%d", a[i]);
		}
		printf("\n");
	}
	printf("\n");
	/*print rot-k-e*/

	/*print ref-k-b*/
	for (int k = 0; k <= N; k++) {
		ref(k, a, b);
		printf("reflection-%d\n", k);
		for (int i = 0; i < N; i++) {
			printf("%d", a[i]);
		}
		printf("\n");
	}
	printf("\n");
	/*print ref-k-e*/

	/*print ref-k -> ror-k+1 -b*/
	for (int k = 0; k < N; k++) {
		refandrot(k,a,b,c);
		printf("reflection-%d then rotation-%d\n",k,k+1);
		for (int i = 0; i < N; i++) {
			printf("%d", a[i]);
		}
		printf("\n");
	}
	printf("\n");
	/*print ref-k -> ror-k+1 -e*/

	/*computer all the arrangement by reflecting then rotating and print out the distinct arrangement of this case-b*/
	int q = 0;
	//printf("ref-5 mean it don't reflect\n\n");
		for (int k = 0; k <=N; k++) {
				for (int j = 1; j <= N; j++) {
				rxry(k, j, a, b, c);
					//printf("arrangment(ref-rot)- %d %d is\n ", k, j);
						for (int w = 0; w < N; w++) {
								d[q][w] = a[w];
								//printf("%d", d[q][w]);
							}
						//printf("\n");
						q++;
				}
		}
	printf("\n");
	for (int a = 0; a < A; a++) {
		for (int b = 0; b < N; b++) {
			e[a][b] = 0;
		}
	}
	int g = 0;
	for (int a = 0; a < A; a++) {
		int z = 0;
		for (int h = 0; h < A; h++) {
			for (int b=0; b < N; b++) {
				if (e[h][b] == d[a][b]) {
					g++;
				}
			}
			if (g == 5) {
				z++;
			}
			g = 0;
		}
		if (z == 0) {
			for (int b = 0; b < N; b++) {
				e[a][b] = d[a][b];
			}
			printf("the distinct arrangment is\n");
			for (int b = 0; b < N; b++) {
				printf("%d",e[a][b]);
			}
			printf("\n");
		}
	}
	/*computer all the arrangement by reflecting then rotating and print out the distinct arrangement of this case-e*/
	printf("This result tell us that all the arrangemrnt by reflecting then rotating can be constructed with once rotation or once reflection");
	system("pause");
	return 0;
}
void rot(int k,int a[],int b[]) {
	for (int i = 0; i < N; i++) {
		a[i] = b[(i + k) % 5];
	}
}
void ref(int k, int a[], int b[]) {
	if (k == N) {
		for (int i = 0; i < N; i++) {
			a[i] = b[i];
		}
	}
	else {
		for (int i = 0; i < N; i++) {
			a[i] = b[(N + k - i + k) % 5];
		}
	}
}
void refandrot(int k, int a[],int b[],int c[]) {
	for (int i = 0; i < N; i++) {
		a[i] = b[(N + k - i + k) % 5];
		c[i] = a[i];
	}
	for (int i = 0; i < N; i++) {
		a[i] = c[(i + k+1) % 5];
	}
}
void rxry(int k, int K, int a[], int b[], int c[]) {
	if (k == N) {
		for (int i = 0; i < N; i++) {
			a[i] = b[i];
			c[i] = b[i];
		}
	}
	else {
		for (int i = 0; i < N; i++) {
			a[i] = b[(N + k - i + k) % 5];
			c[i] = a[i];
		}
	}
	for (int i = 0; i < N; i++) {
		a[i] = c[(i + K) % 5];
	}
}