A square pattern of size N x N (1 <= N <= 10) black and white square tiles is transformed into another square pattern. Write a program that will recognize the minimum transformation that has been applied to the original pattern given the following list of possible transformations:
- #1: 90 Degree Rotation: The pattern was rotated clockwise 90 degrees.
- #2: 180 Degree Rotation: The pattern was rotated clockwise 180 degrees.
- #3: 270 Degree Rotation: The pattern was rotated clockwise 270 degrees.
- #4: Reflection: The pattern was reflected horizontally (turned into a mirror image of itself by reflecting around a vertical line in the middle of the image).
- #5: Combination: The pattern was reflected horizontally and then subjected to one of the rotations (#1-#3).
- #6: No Change: The original pattern was not changed.
- #7: Invalid Transformation: The new pattern was not obtained by any of the above methods.
In the case that more than one transform could have been used, choose the one with the minimum number above.
PROGRAM NAME: transform
INPUT FORMAT
| Line 1: | A single integer, N |
| Line 2..N+1: | N lines of N characters (each either `@' or `-'); this is the square before transformation |
| Line N+2..2*N+1: | N lines of N characters (each either `@' or `-'); this is the square after transformation |
SAMPLE INPUT (file transform.in)
3 @-@ --- @@- @-@ @-- --@
OUTPUT FORMAT
A single line containing the the number from 1 through 7 (described above) that categorizes the transformation required to change from the `before' representation to the `after' representation.
SAMPLE OUTPUT (file transform.out)
1
题解：a数组是转换前的状态，b数组是转换后的状态。
只要枚举这七种情况即可
顺时针转90度：b[i][j]=a[n-j-1][i]；
顺时针转180度：b[i][j]=a[n-i-1][n-j-1];
顺时针转270度：b[i][j]=a[j][n-i-1]；
水平翻转：b[i][j]=a[i][n-j-1];
第五种情况的话，先把b水平翻转一下，然后再判断是否符合1-3中的一种。
不改变：b[i][j]=a[i][j];
如果以上六种情况都不符合，那就直接输出7就OK了。。。
View Code
1 /* 2 ID:spcjv51 3 PROG:transform 4 LANG:C 5 */ 6 #include<stdio.h> 7 #include<string.h> 8 #define maxn 11 9 char a[maxn][maxn],b[maxn][maxn],c[maxn][maxn]; 10 int n; 11 int trans1() 12 { 13 int i,j; 14 for(i=0; i<n; i++) 15 for(j=0; j<n; j++) 16 if(b[i][j]!=a[n-j-1][i]) return 0; 17 return 1; 18 } 19 int trans2() 20 { 21 int i,j; 22 for(i=0; i<n; i++) 23 for(j=0; j<n; j++) 24 if(b[i][j]!=a[n-i-1][n-j-1]) return 0; 25 return 1; 26 27 } 28 int trans3() 29 { 30 int i,j; 31 for(i=0; i<n; i++) 32 for(j=0; j<n; j++) 33 if(b[i][j]!=a[j][n-i-1]) return 0; 34 return 1; 35 } 36 int trans4() 37 { 38 int i,j; 39 for(i=0; i<n; i++) 40 for(j=0; j<n; j++) 41 if(b[i][j]!=a[i][n-j-1]) return 0; 42 return 1; 43 } 44 int trans6() 45 { 46 int i,j; 47 for(i=0; i<n; i++) 48 for(j=0; j<n; j++) 49 if(b[i][j]!=a[i][j]) return 0; 50 return 1; 51 } 52 53 int main(void) 54 { 55 freopen("transform.in","r",stdin); 56 freopen("transform.out","w",stdout); 57 int i,j,m,ans,flag; 58 scanf("%d",&n); 59 for(i=0; i<n; i++) 60 scanf("%s",a[i]); 61 for(i=0; i<n; i++) 62 scanf("%s",b[i]); 63 flag=1; 64 if(flag&&trans1()) 65 { 66 printf("1\n"); 67 flag=0; 68 } 69 if(flag&&trans2()) 70 { 71 printf("2\n"); 72 flag=0; 73 } 74 if(flag&&trans3()) 75 { 76 printf("3\n"); 77 flag=0; 78 } 79 if(flag&&trans4()) 80 { 81 printf("4\n"); 82 flag=0; 83 } 84 if(flag) 85 { 86 int i,j; 87 for(i=0; i<n; i++) 88 for(j=0; j<n; j++) 89 c[i][j]=b[i][n-j-1]; 90 for(i=0; i<n; i++) 91 for(j=0;j<n;j++) 92 b[i][j]=c[i][j]; 93 if(trans1()||trans2()||trans3()) 94 { 95 printf("5\n"); 96 flag=0; 97 } 98 } 99 100 if(flag&&trans6()) 101 { 102 printf("6\n"); 103 flag=0; 104 } 105 if(flag) printf("7\n"); 106 return 0; 107 }