INPUT > OUTPUT INPUT > > OUTPUT
4 4
1 2 3 4 1 1 1 1 11 12 13 14 41 31 21 11
1 2 3 4 2 2 2 2 21 22 23 24 42 32 22 12
1 2 3 4 3 3 3 3 31 32 33 34 43 33 23 13
1 2 3 4 4 4 4 4 41 42 43 44 44 34 24 14
#include<stdio.h>
void rotate(int I[10][10],int O[10][10],int j,int i)
{
int k,l,temp;
for(k=0;k<j;k++)
{
temp=I[k+i][i]; // copy of Left in temp
O[k+i][i]=I[i+j-1][i+k]; // copy bottom to left
O[i+j-1][i+k]=I[i+j-k-1][i+j-1]; // copy right to bottom
O[i+j-k-1][i+j-1]=I[i][i+j-k-1]; // copy top to right
O[i][i+j-k-1]=temp; // copy temp(left) to top
}
}// end of rotate
int main()
{
int I[10][10],O[10][10],i,j,N;
scanf("%d",&N);
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
scanf("%d",&I[i][j]);
}
i=0;
j=N;
while(j>0)
{
rotate(I,O,j,i); // i is base, j is size
j=j-2;
i++;
}
printf("\n\n");
for(i=0;i<N;i++)
{
printf("\n");
for(j=0;j<N;j++)
printf("%d ",O[i][j]);
}
return 0;
}
the problem states it it needs to be done in place.
ReplyDeleteit seems to me you are using two matrices?
#include
ReplyDeleteusing namespace std;
int a[5][5] = { {0,0,0,0,0},
{1,1,1,1,1},
{1,2,3,4,5},
{2,2,2,2,2},
{3,3,3,3,3}
};
void print()
{
for(int i = 0;i<5; ++i) {
for(int j = 0;j<5; ++j)
cout << a[i][j] << " ";
cout<<endl;
}
cout<<endl;
}
void flip() {
for(int i = 0; i<=2; ++i) {
for(int j = i;j<4-i; ++j) {
int tmp = a[i][j];
a[i][j] = a[j][4-i];
a[j][4-i] = a[4-i][4-j];
a[4-i][4-j] = a[4 -j][i];
a[4-j][i] = tmp;
}
}
}
int main()
{
print();
flip();
print();
flip();
print();
return 0;
}
flip has 2 important points
ReplyDelete1. calculating the indices for 4 nodes from left, right, bott, top sizdes that are going to be switched each time.
2. for loops stop conditions. basically what happens after first run (i=0, j=0...4), we switch all the side cells. so next time we need to skip all outer cells and go further in.
Take a look, run an example, try to see how it works.